Please compute $int_0^{infty}e^{-(x+ai)^2}dx$, where, $i^2=-1$.Are Complex Substitutions Legal in...

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Please compute $int_0^{infty}e^{-(x+ai)^2}dx$, where, $i^2=-1$.


Are Complex Substitutions Legal in Integration?How to compute this integral of Bessel functions?A Hankel Transform Integral: $int_0^inftyfrac{1}{k^2-k_p^2}J_0left(krhoright);k,dk$Is the Hankel Transform a Hankel OperatorRadial Green's functionsolution for integral $int_0^{infty} frac{k}{k^3-a}J_0left(k , rright) dk $ involving Bessel function (Hankel transform)Integration of zeroth order Bessel function of first kind divided by a polynomialIntegral of product of 3 bessel functionsHankel Transform of Modified GaussianFinding the Fourier-Bessel Series For $f(x) = x$, $0 < x < 2$, With Respect to the Orthogonal Set ${ J_1 (k_n(x)) }$What is the Fourier transform of the 2 dimensional airy function?













1












$begingroup$


I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function:
$$
int_0^{infty} !int_0^{infty} ! left( Are^{-(r^2+z^2)/delta^2}right)J_1(k_r r) e^{-ik_z z} r , mathrm{d}r, mathrm{d}z
$$
Trying to compute the integral in $z$ I got
$$int_0^{infty}e^{-left(zeta+i k_z delta/2right)^2},mathrm{d}zeta.$$
Any ideas on how to attempt this?



Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.






contour-integration fourier-transform bessel-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
    $endgroup$
    – Michael Seifert
    Mar 18 at 20:21












  • $begingroup$
    Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
    $endgroup$
    – Teebro Prokash
    Mar 18 at 20:51












  • $begingroup$
    With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
    $endgroup$
    – Michael Seifert
    Mar 18 at 22:06


















1












$begingroup$


I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function:
$$
int_0^{infty} !int_0^{infty} ! left( Are^{-(r^2+z^2)/delta^2}right)J_1(k_r r) e^{-ik_z z} r , mathrm{d}r, mathrm{d}z
$$
Trying to compute the integral in $z$ I got
$$int_0^{infty}e^{-left(zeta+i k_z delta/2right)^2},mathrm{d}zeta.$$
Any ideas on how to attempt this?



Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.






contour-integration fourier-transform bessel-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
    $endgroup$
    – Michael Seifert
    Mar 18 at 20:21












  • $begingroup$
    Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
    $endgroup$
    – Teebro Prokash
    Mar 18 at 20:51












  • $begingroup$
    With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
    $endgroup$
    – Michael Seifert
    Mar 18 at 22:06
















1












1








1





$begingroup$


I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function:
$$
int_0^{infty} !int_0^{infty} ! left( Are^{-(r^2+z^2)/delta^2}right)J_1(k_r r) e^{-ik_z z} r , mathrm{d}r, mathrm{d}z
$$
Trying to compute the integral in $z$ I got
$$int_0^{infty}e^{-left(zeta+i k_z delta/2right)^2},mathrm{d}zeta.$$
Any ideas on how to attempt this?



Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.






contour-integration fourier-transform bessel-functions







share|cite|improve this question











$endgroup$




I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function:
$$
int_0^{infty} !int_0^{infty} ! left( Are^{-(r^2+z^2)/delta^2}right)J_1(k_r r) e^{-ik_z z} r , mathrm{d}r, mathrm{d}z
$$
Trying to compute the integral in $z$ I got
$$int_0^{infty}e^{-left(zeta+i k_z delta/2right)^2},mathrm{d}zeta.$$
Any ideas on how to attempt this?



Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.






contour-integration fourier-transform bessel-functions




contour-integration fourier-transform bessel-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 20:50







Teebro Prokash

















asked Mar 18 at 19:32









Teebro ProkashTeebro Prokash

8010




8010












  • $begingroup$
    For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
    $endgroup$
    – Michael Seifert
    Mar 18 at 20:21












  • $begingroup$
    Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
    $endgroup$
    – Teebro Prokash
    Mar 18 at 20:51












  • $begingroup$
    With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
    $endgroup$
    – Michael Seifert
    Mar 18 at 22:06




















  • $begingroup$
    For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
    $endgroup$
    – Michael Seifert
    Mar 18 at 20:21












  • $begingroup$
    Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
    $endgroup$
    – Teebro Prokash
    Mar 18 at 20:51












  • $begingroup$
    With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
    $endgroup$
    – Michael Seifert
    Mar 18 at 22:06


















$begingroup$
For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
$endgroup$
– Michael Seifert
Mar 18 at 20:21






$begingroup$
For the $r$ integral: from the various recurrence relations for the Bessel functions one can show that $$int x^2 J_1(x) dx = - x^2 J_0(x) + 2 int x J_0(x) dx = - x^2 J_0(x) + 2 x J_1(x).$$However, this antiderivative does not approach a finite limit as $x to infty$, which implies that your radial integral does not converge.
$endgroup$
– Michael Seifert
Mar 18 at 20:21














$begingroup$
Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
$endgroup$
– Teebro Prokash
Mar 18 at 20:51






$begingroup$
Sorry, there is also a $e^{-r^2/delta^2}$. My bad.
$endgroup$
– Teebro Prokash
Mar 18 at 20:51














$begingroup$
With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
$endgroup$
– Michael Seifert
Mar 18 at 22:06






$begingroup$
With that change, Mathematica gives your answer for the radial integral as $frac{delta^4}{4} e^{-delta^2 k_r^2/4}$, but damned if I know how to show it.
$endgroup$
– Michael Seifert
Mar 18 at 22:06












2 Answers
2






active

oldest

votes


















2












$begingroup$

Consider a closed contour $gamma$ in the complex plane consisting of four straight line segments:




  1. From $z = 0$ to $z = R in mathbb{R}$;

  2. From $z = R$ to $z = R + i a$;

  3. From $z = R + i a$ to $z = i a$;

  4. From $z = ia$ to $z = 0$.


Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $gamma$ is zero:
$$
oint f(z) , dz = int_{gamma_1} f(z) , dz + int_{gamma_2} f(z) , dz +int_{gamma_3} f(z) , dz +int_{gamma_4} f(z) , dz = 0.
$$
Taking each of these four integrals in turn, and looking at their limit as $R to infty$:




  1. We have $$
    int_{gamma_1} f(z) , dz = int_0^R e^{-x^2} , dx,
    $$
    and in the limit $R to infty$, this approaches $sqrt{pi}/2$.


  2. We have
    $$
    int_{gamma_2} f(z) , dz = i int_0^a e^{-(R + iy)^2}, dy = i e^{-R^2} int_0^a e^{2 i R y} e^{y^2} , dy.
    $$
    The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R to infty$, the integrand over $gamma_2$ will vanish.


  3. We have
    $$
    int_{gamma_3} f(z) , dz = int_R^0 e^{-(x + ia) ^2} , dx = - int_0^R e^{-(x + ia) ^2} , dx,
    $$
    and so as $R to infty$ this approaches the negative of the integral we're looking for.


  4. We have
    $$
    int_{gamma_4} f(z) , dz = i int_a^0 e^{-(iy)^2}, dy = - i int_0^a e^{y^2} , dy.
    $$



Thus, putting all four integrals together and taking the $R to infty$ limit, we conclude that
$$
frac{sqrt{pi}}{2} - int_0^infty e^{-(x + ia) ^2} , dx - i int_0^a e^{y^2} , dy = 0 \
boxed{ int_0^infty e^{-(x + ia) ^2} = frac{sqrt{pi}}{2} - i int_0^a e^{y^2} dy.}
$$



The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $mathrm{erf}(z)$. In this case, we have
$$
int_0^a e^{y^2} dy = - frac{i sqrt{pi}}{2} mathrm{erf}(i a).
$$
Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $int_{ai}^infty e^{-u^2}du$ which can be written in terms of the "error function".






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
      $endgroup$
      – Winther
      Mar 18 at 20:00












    • $begingroup$
      I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
      $endgroup$
      – Teebro Prokash
      Mar 18 at 20:10










    • $begingroup$
      I completely agree with @Winther
      $endgroup$
      – Jean Marie
      Mar 18 at 21:05












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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Consider a closed contour $gamma$ in the complex plane consisting of four straight line segments:




    1. From $z = 0$ to $z = R in mathbb{R}$;

    2. From $z = R$ to $z = R + i a$;

    3. From $z = R + i a$ to $z = i a$;

    4. From $z = ia$ to $z = 0$.


    Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $gamma$ is zero:
    $$
    oint f(z) , dz = int_{gamma_1} f(z) , dz + int_{gamma_2} f(z) , dz +int_{gamma_3} f(z) , dz +int_{gamma_4} f(z) , dz = 0.
    $$
    Taking each of these four integrals in turn, and looking at their limit as $R to infty$:




    1. We have $$
      int_{gamma_1} f(z) , dz = int_0^R e^{-x^2} , dx,
      $$
      and in the limit $R to infty$, this approaches $sqrt{pi}/2$.


    2. We have
      $$
      int_{gamma_2} f(z) , dz = i int_0^a e^{-(R + iy)^2}, dy = i e^{-R^2} int_0^a e^{2 i R y} e^{y^2} , dy.
      $$
      The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R to infty$, the integrand over $gamma_2$ will vanish.


    3. We have
      $$
      int_{gamma_3} f(z) , dz = int_R^0 e^{-(x + ia) ^2} , dx = - int_0^R e^{-(x + ia) ^2} , dx,
      $$
      and so as $R to infty$ this approaches the negative of the integral we're looking for.


    4. We have
      $$
      int_{gamma_4} f(z) , dz = i int_a^0 e^{-(iy)^2}, dy = - i int_0^a e^{y^2} , dy.
      $$



    Thus, putting all four integrals together and taking the $R to infty$ limit, we conclude that
    $$
    frac{sqrt{pi}}{2} - int_0^infty e^{-(x + ia) ^2} , dx - i int_0^a e^{y^2} , dy = 0 \
    boxed{ int_0^infty e^{-(x + ia) ^2} = frac{sqrt{pi}}{2} - i int_0^a e^{y^2} dy.}
    $$



    The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $mathrm{erf}(z)$. In this case, we have
    $$
    int_0^a e^{y^2} dy = - frac{i sqrt{pi}}{2} mathrm{erf}(i a).
    $$
    Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Consider a closed contour $gamma$ in the complex plane consisting of four straight line segments:




      1. From $z = 0$ to $z = R in mathbb{R}$;

      2. From $z = R$ to $z = R + i a$;

      3. From $z = R + i a$ to $z = i a$;

      4. From $z = ia$ to $z = 0$.


      Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $gamma$ is zero:
      $$
      oint f(z) , dz = int_{gamma_1} f(z) , dz + int_{gamma_2} f(z) , dz +int_{gamma_3} f(z) , dz +int_{gamma_4} f(z) , dz = 0.
      $$
      Taking each of these four integrals in turn, and looking at their limit as $R to infty$:




      1. We have $$
        int_{gamma_1} f(z) , dz = int_0^R e^{-x^2} , dx,
        $$
        and in the limit $R to infty$, this approaches $sqrt{pi}/2$.


      2. We have
        $$
        int_{gamma_2} f(z) , dz = i int_0^a e^{-(R + iy)^2}, dy = i e^{-R^2} int_0^a e^{2 i R y} e^{y^2} , dy.
        $$
        The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R to infty$, the integrand over $gamma_2$ will vanish.


      3. We have
        $$
        int_{gamma_3} f(z) , dz = int_R^0 e^{-(x + ia) ^2} , dx = - int_0^R e^{-(x + ia) ^2} , dx,
        $$
        and so as $R to infty$ this approaches the negative of the integral we're looking for.


      4. We have
        $$
        int_{gamma_4} f(z) , dz = i int_a^0 e^{-(iy)^2}, dy = - i int_0^a e^{y^2} , dy.
        $$



      Thus, putting all four integrals together and taking the $R to infty$ limit, we conclude that
      $$
      frac{sqrt{pi}}{2} - int_0^infty e^{-(x + ia) ^2} , dx - i int_0^a e^{y^2} , dy = 0 \
      boxed{ int_0^infty e^{-(x + ia) ^2} = frac{sqrt{pi}}{2} - i int_0^a e^{y^2} dy.}
      $$



      The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $mathrm{erf}(z)$. In this case, we have
      $$
      int_0^a e^{y^2} dy = - frac{i sqrt{pi}}{2} mathrm{erf}(i a).
      $$
      Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Consider a closed contour $gamma$ in the complex plane consisting of four straight line segments:




        1. From $z = 0$ to $z = R in mathbb{R}$;

        2. From $z = R$ to $z = R + i a$;

        3. From $z = R + i a$ to $z = i a$;

        4. From $z = ia$ to $z = 0$.


        Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $gamma$ is zero:
        $$
        oint f(z) , dz = int_{gamma_1} f(z) , dz + int_{gamma_2} f(z) , dz +int_{gamma_3} f(z) , dz +int_{gamma_4} f(z) , dz = 0.
        $$
        Taking each of these four integrals in turn, and looking at their limit as $R to infty$:




        1. We have $$
          int_{gamma_1} f(z) , dz = int_0^R e^{-x^2} , dx,
          $$
          and in the limit $R to infty$, this approaches $sqrt{pi}/2$.


        2. We have
          $$
          int_{gamma_2} f(z) , dz = i int_0^a e^{-(R + iy)^2}, dy = i e^{-R^2} int_0^a e^{2 i R y} e^{y^2} , dy.
          $$
          The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R to infty$, the integrand over $gamma_2$ will vanish.


        3. We have
          $$
          int_{gamma_3} f(z) , dz = int_R^0 e^{-(x + ia) ^2} , dx = - int_0^R e^{-(x + ia) ^2} , dx,
          $$
          and so as $R to infty$ this approaches the negative of the integral we're looking for.


        4. We have
          $$
          int_{gamma_4} f(z) , dz = i int_a^0 e^{-(iy)^2}, dy = - i int_0^a e^{y^2} , dy.
          $$



        Thus, putting all four integrals together and taking the $R to infty$ limit, we conclude that
        $$
        frac{sqrt{pi}}{2} - int_0^infty e^{-(x + ia) ^2} , dx - i int_0^a e^{y^2} , dy = 0 \
        boxed{ int_0^infty e^{-(x + ia) ^2} = frac{sqrt{pi}}{2} - i int_0^a e^{y^2} dy.}
        $$



        The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $mathrm{erf}(z)$. In this case, we have
        $$
        int_0^a e^{y^2} dy = - frac{i sqrt{pi}}{2} mathrm{erf}(i a).
        $$
        Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.






        share|cite|improve this answer









        $endgroup$



        Consider a closed contour $gamma$ in the complex plane consisting of four straight line segments:




        1. From $z = 0$ to $z = R in mathbb{R}$;

        2. From $z = R$ to $z = R + i a$;

        3. From $z = R + i a$ to $z = i a$;

        4. From $z = ia$ to $z = 0$.


        Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $gamma$ is zero:
        $$
        oint f(z) , dz = int_{gamma_1} f(z) , dz + int_{gamma_2} f(z) , dz +int_{gamma_3} f(z) , dz +int_{gamma_4} f(z) , dz = 0.
        $$
        Taking each of these four integrals in turn, and looking at their limit as $R to infty$:




        1. We have $$
          int_{gamma_1} f(z) , dz = int_0^R e^{-x^2} , dx,
          $$
          and in the limit $R to infty$, this approaches $sqrt{pi}/2$.


        2. We have
          $$
          int_{gamma_2} f(z) , dz = i int_0^a e^{-(R + iy)^2}, dy = i e^{-R^2} int_0^a e^{2 i R y} e^{y^2} , dy.
          $$
          The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R to infty$, the integrand over $gamma_2$ will vanish.


        3. We have
          $$
          int_{gamma_3} f(z) , dz = int_R^0 e^{-(x + ia) ^2} , dx = - int_0^R e^{-(x + ia) ^2} , dx,
          $$
          and so as $R to infty$ this approaches the negative of the integral we're looking for.


        4. We have
          $$
          int_{gamma_4} f(z) , dz = i int_a^0 e^{-(iy)^2}, dy = - i int_0^a e^{y^2} , dy.
          $$



        Thus, putting all four integrals together and taking the $R to infty$ limit, we conclude that
        $$
        frac{sqrt{pi}}{2} - int_0^infty e^{-(x + ia) ^2} , dx - i int_0^a e^{y^2} , dy = 0 \
        boxed{ int_0^infty e^{-(x + ia) ^2} = frac{sqrt{pi}}{2} - i int_0^a e^{y^2} dy.}
        $$



        The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $mathrm{erf}(z)$. In this case, we have
        $$
        int_0^a e^{y^2} dy = - frac{i sqrt{pi}}{2} mathrm{erf}(i a).
        $$
        Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 18 at 22:00









        Michael SeifertMichael Seifert

        5,172626




        5,172626























            -1












            $begingroup$

            Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $int_{ai}^infty e^{-u^2}du$ which can be written in terms of the "error function".






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
              $endgroup$
              – Winther
              Mar 18 at 20:00












            • $begingroup$
              I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
              $endgroup$
              – Teebro Prokash
              Mar 18 at 20:10










            • $begingroup$
              I completely agree with @Winther
              $endgroup$
              – Jean Marie
              Mar 18 at 21:05
















            -1












            $begingroup$

            Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $int_{ai}^infty e^{-u^2}du$ which can be written in terms of the "error function".






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
              $endgroup$
              – Winther
              Mar 18 at 20:00












            • $begingroup$
              I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
              $endgroup$
              – Teebro Prokash
              Mar 18 at 20:10










            • $begingroup$
              I completely agree with @Winther
              $endgroup$
              – Jean Marie
              Mar 18 at 21:05














            -1












            -1








            -1





            $begingroup$

            Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $int_{ai}^infty e^{-u^2}du$ which can be written in terms of the "error function".






            share|cite|improve this answer









            $endgroup$



            Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $int_{ai}^infty e^{-u^2}du$ which can be written in terms of the "error function".







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 18 at 19:56









            user247327user247327

            11.6k1516




            11.6k1516








            • 2




              $begingroup$
              See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
              $endgroup$
              – Winther
              Mar 18 at 20:00












            • $begingroup$
              I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
              $endgroup$
              – Teebro Prokash
              Mar 18 at 20:10










            • $begingroup$
              I completely agree with @Winther
              $endgroup$
              – Jean Marie
              Mar 18 at 21:05














            • 2




              $begingroup$
              See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
              $endgroup$
              – Winther
              Mar 18 at 20:00












            • $begingroup$
              I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
              $endgroup$
              – Teebro Prokash
              Mar 18 at 20:10










            • $begingroup$
              I completely agree with @Winther
              $endgroup$
              – Jean Marie
              Mar 18 at 21:05








            2




            2




            $begingroup$
            See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
            $endgroup$
            – Winther
            Mar 18 at 20:00






            $begingroup$
            See Are Complex Substitutions Legal in Integration?. Doing this you are pretending $i$ is a real number. This procedure most likely gives the correct result, but it's not correct.
            $endgroup$
            – Winther
            Mar 18 at 20:00














            $begingroup$
            I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
            $endgroup$
            – Teebro Prokash
            Mar 18 at 20:10




            $begingroup$
            I'm somewhat more interested to know what this evaluates to. Can you provide with a formula for error function evaluated at complex (esp. purely imaginary) values?
            $endgroup$
            – Teebro Prokash
            Mar 18 at 20:10












            $begingroup$
            I completely agree with @Winther
            $endgroup$
            – Jean Marie
            Mar 18 at 21:05




            $begingroup$
            I completely agree with @Winther
            $endgroup$
            – Jean Marie
            Mar 18 at 21:05


















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