Evaluate $int_{-infty}^{infty}frac{dx}{(1+x^2)^z}$ where $Re(z)>frac{1}{2}$ Announcing the...

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Evaluate $int_{-infty}^{infty}frac{dx}{(1+x^2)^z}$ where $Re(z)>frac{1}{2}$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integral evaluation $int_{-infty}^{infty}frac{cos (ax)}{pi (1+x^2)}dx$Evaluating the complex integral $int_{0}^{infty} frac{sin(x+i)}{x+ i } , mathscr{d}x$Show that $int_{0}^{infty} frac{cosh(ax)}{cosh(pi x)} dx=frac{1}{2}sec(frac{a}{2})$ using Residue CalculusShow that $int_{C} frac{e^{az}}{sin(pi z)} ,dz rightarrow 0 quad as quad R rightarrow infty$Contour integral $int_{-infty}^{infty}e^{-iax}/(-b+cos(x))mathrm dx$ with $a>0$ and $0<b<1$Complex integration of $int_{-infty}^{infty}frac{dx}{(x^2+2)^3}$complex integration $frac{1-|a|^{2}}{pi}int_{L}frac{|dz|}{{|z+a|}^{2}}$Evaluate the following: $int_{0}^{infty}frac{sin(ax)}{sinh(x)}dx$ and $int_{0}^{infty}frac{xcos(ax)}{sinh(x)}dx$ using rectangular contourEvaluate $int_{-infty }^{infty } frac{sin^2x}{x^2} , dx$Evaluating the complex integral $int_{0}^{infty} frac {sin (ln x) dx }{x^2 + 4} $












1












$begingroup$


I am dealing with the following complex integration:



$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$



But I do not know how exactly to deal with the $z$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
    $endgroup$
    – Jack D'Aurizio
    Mar 23 at 13:18










  • $begingroup$
    @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
    $endgroup$
    – Mark Viola
    Mar 29 at 16:28
















1












$begingroup$


I am dealing with the following complex integration:



$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$



But I do not know how exactly to deal with the $z$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
    $endgroup$
    – Jack D'Aurizio
    Mar 23 at 13:18










  • $begingroup$
    @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
    $endgroup$
    – Mark Viola
    Mar 29 at 16:28














1












1








1





$begingroup$


I am dealing with the following complex integration:



$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$



But I do not know how exactly to deal with the $z$.










share|cite|improve this question











$endgroup$




I am dealing with the following complex integration:



$$int_{-infty}^{infty}frac{dx}{(1+x^2)^z} quadtext{ where }quadRe(z)>frac{1}{2}$$



But I do not know how exactly to deal with the $z$.







complex-analysis complex-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 13:52









StubbornAtom

6,42731440




6,42731440










asked Mar 23 at 12:44









thdrthdr

62




62












  • $begingroup$
    By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
    $endgroup$
    – Jack D'Aurizio
    Mar 23 at 13:18










  • $begingroup$
    @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
    $endgroup$
    – Mark Viola
    Mar 29 at 16:28


















  • $begingroup$
    By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
    $endgroup$
    – Jack D'Aurizio
    Mar 23 at 13:18










  • $begingroup$
    @thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
    $endgroup$
    – Mark Viola
    Mar 29 at 16:28
















$begingroup$
By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
$endgroup$
– Jack D'Aurizio
Mar 23 at 13:18




$begingroup$
By exploiting symmetry and enforcing the substitution $x=tantheta$ you immediately get something manageable through Euler's Beta function.
$endgroup$
– Jack D'Aurizio
Mar 23 at 13:18












$begingroup$
@thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 29 at 16:28




$begingroup$
@thdr Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 29 at 16:28










2 Answers
2






active

oldest

votes


















1












$begingroup$

Using Schwinger parametrization one can write
$$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
Then switching the order of integration we have
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
$$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
So the integral becomes
$$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
$$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
Giving the final answer of
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $f(z)$ be given by



    $$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$



    Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$



    $$begin{align}
    f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
    &=B(1/2,z-1/2)\\
    &=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
      $endgroup$
      – Mark Viola
      Mar 29 at 16:29












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Using Schwinger parametrization one can write
    $$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
    Then switching the order of integration we have
    $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
    The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
    $$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
    So the integral becomes
    $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
    and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
    $$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
    Giving the final answer of
    $$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Using Schwinger parametrization one can write
      $$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
      Then switching the order of integration we have
      $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
      The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
      $$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
      So the integral becomes
      $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
      and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
      $$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
      Giving the final answer of
      $$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Using Schwinger parametrization one can write
        $$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
        Then switching the order of integration we have
        $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
        The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
        $$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
        So the integral becomes
        $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
        and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
        $$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
        Giving the final answer of
        $$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$






        share|cite|improve this answer









        $endgroup$



        Using Schwinger parametrization one can write
        $$int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=frac{1}{Gamma(z)}int_{-infty}^{infty}int_0^infty t^{z-1}e^{-t(1+x^2)}dtdx$$
        Then switching the order of integration we have
        $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}int_{-infty}^infty e^{-tx^2}dxdt$$
        The second integral is a Gaussian integral and can be found by using the substitution $s=sqrt{t}xrightarrow ds=sqrt{t}dx$ which gives
        $$int_{-infty}^infty e^{-tx^2}dx=frac1{sqrt{t}}int_{-infty}^infty e^{-s^2}ds=sqrt{frac{pi}t}$$
        So the integral becomes
        $$frac{1}{Gamma(z)}int_0^{infty}t^{z-1}e^{-t}sqrt{frac{pi}t}dt=frac{sqrt{pi}}{Gamma(z)}int_0^{infty}t^{z-1.5}e^{-t}dt$$
        and then the integral left is the Gamma function - $Gamma(z-0.5)$ - because
        $$int_0^infty t^{z-1}e^{-t}dt=Gamma(z)$$
        Giving the final answer of
        $$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 23 at 13:08









        Peter ForemanPeter Foreman

        7,2411319




        7,2411319























            0












            $begingroup$

            Let $f(z)$ be given by



            $$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$



            Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$



            $$begin{align}
            f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
            &=B(1/2,z-1/2)\\
            &=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
            end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
              $endgroup$
              – Mark Viola
              Mar 29 at 16:29
















            0












            $begingroup$

            Let $f(z)$ be given by



            $$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$



            Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$



            $$begin{align}
            f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
            &=B(1/2,z-1/2)\\
            &=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
            end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
              $endgroup$
              – Mark Viola
              Mar 29 at 16:29














            0












            0








            0





            $begingroup$

            Let $f(z)$ be given by



            $$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$



            Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$



            $$begin{align}
            f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
            &=B(1/2,z-1/2)\\
            &=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
            end{align}$$






            share|cite|improve this answer









            $endgroup$



            Let $f(z)$ be given by



            $$f(z) =int_0^infty frac1{(1+x^2)^z},dx$$



            Since the integrand is an even function of $x$ we have after enforcing the substitution $xmapsto sqrt x$



            $$begin{align}
            f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
            &=B(1/2,z-1/2)\\
            &=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
            end{align}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 23 at 14:51









            Mark ViolaMark Viola

            134k1278177




            134k1278177












            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
              $endgroup$
              – Mark Viola
              Mar 29 at 16:29


















            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
              $endgroup$
              – Mark Viola
              Mar 29 at 16:29
















            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
            $endgroup$
            – Mark Viola
            Mar 29 at 16:29




            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
            $endgroup$
            – Mark Viola
            Mar 29 at 16:29


















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