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} $
$begingroup$
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 $z$.
complex-analysis complex-integration
$endgroup$
add a comment |
$begingroup$
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 $z$.
complex-analysis complex-integration
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $z$.
complex-analysis complex-integration
$endgroup$
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
complex-analysis complex-integration
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$
$endgroup$
add a comment |
$begingroup$
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 $xmapsto sqrt x$
$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$
$endgroup$
add a comment |
$begingroup$
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
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$
$endgroup$
add a comment |
$begingroup$
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
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$
$endgroup$
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
$$boxed{int_{-infty}^{infty}frac{1}{(1+x^2)^z}dx=sqrt{pi}frac{Gamma(z-0.5)}{Gamma(z)}}$$
answered Mar 23 at 13:08
Peter ForemanPeter Foreman
7,2411319
7,2411319
add a comment |
add a comment |
$begingroup$
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 $xmapsto sqrt x$
$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xmapsto sqrt x$
$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xmapsto sqrt x$
$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$
$endgroup$
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 $xmapsto sqrt x$
$$begin{align}
f(z)&=int_0^infty frac{x^{-1/2}}{(1+x)^z},dx\\
&=B(1/2,z-1/2)\\
&=sqrt pi ,frac{Gamma(z-1/2)}{Gamma(z)}
end{align}$$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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