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Compute $int_{0}^{+infty} frac{sin x}{sqrt{x}} dx$ using Gamma function
$begingroup$
I want to compute
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx$$
using Gamma function.
I know that by change of variable, $y=sqrt{x}$, one gets
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx=2int_{0}^{+infty}sin y^2 dy=frac{sqrt{2pi}}{2}$$
by Fresnel's integral.
I try it by considering this:
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx$$
It converges for both real and imaginary part using Dirichlet test, and $0$ is not a problem here. Let the square root take the pricipal branch where $sqrt{1}=1$. Let $y=-ix$, then
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx=sqrt{i}int_{0}^{-iinfty}y^{-frac{1}{2}}e^{-y} dy=(frac{sqrt{2}}{2} + frac{sqrt{2}}{2}i)Gamma(frac{1}{2})=frac{sqrt{2pi}}{2} + frac{sqrt{2pi}}{2}i$$
And it coincides with the final answer!
My problem is, suppose $L$ is a ray starting from $0$ and has an angle $phi$ with the $x$-axis, and let $phiin(0,2pi)$.
I want to argue that (maybe it is incorrect though)
$$Gamma(z)=int_{L}t^{z-1}e^{-t} dt$$
I firstly know that it converges when $Re(z)>0$. Choose a contour like sector, let $L_1={z=x+iy:y=0, r<x<R}$, $L_2={z=Re^{itheta}:0<theta<phi}$, $L_3={z=xe^{iphi}:r<x<R}$ and $L_4={z=re^{itheta}:0<theta<phi}$, where $r<R$ and the contour is counterclockwise. By Cauchy theorem we have the contour integral should be $0$.
Easy to see that (let $z=x+iy$)
$$
lim_{rto0+, Rto+infty}int_{L_1}t^{z-1}e^{-t} dt=Gamma(z)
$$
$$
|int_{L_4}t^{z-1}e^{-t} dt|=|int_{phi}^{0} e^{-re^{itheta}} (re^{itheta})^{z-1}ire^{itheta} dtheta| leq int_{0}^{phi} e^{-rcostheta} |r^{z}e^{itheta(z-1)}| dtheta=int_{0}^{phi} e^{-rcostheta} r^{x}e^{-theta y} dtheta to 0, r to 0+
$$
But when considering $L_2$:
$$
|int_{L_2}t^{z-1}e^{-t} dt|leqint_{0}^{phi} e^{-Rcostheta} R^{x}e^{-theta y} dtheta
$$
when for example, $frac{3pi}{2}>phi>frac{pi}{2}$, we have $costheta<0$ and I failed to prove the above integral goes to zero when $Rto +infty$.
Is my usage of Gamma function to compute the original integral a coincidence to get the correct result, or there is a way to prove my argument?
Thank you so much!
complex-analysis gamma-function
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
$endgroup$
add a comment |
$begingroup$
I want to compute
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx$$
using Gamma function.
I know that by change of variable, $y=sqrt{x}$, one gets
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx=2int_{0}^{+infty}sin y^2 dy=frac{sqrt{2pi}}{2}$$
by Fresnel's integral.
I try it by considering this:
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx$$
It converges for both real and imaginary part using Dirichlet test, and $0$ is not a problem here. Let the square root take the pricipal branch where $sqrt{1}=1$. Let $y=-ix$, then
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx=sqrt{i}int_{0}^{-iinfty}y^{-frac{1}{2}}e^{-y} dy=(frac{sqrt{2}}{2} + frac{sqrt{2}}{2}i)Gamma(frac{1}{2})=frac{sqrt{2pi}}{2} + frac{sqrt{2pi}}{2}i$$
And it coincides with the final answer!
My problem is, suppose $L$ is a ray starting from $0$ and has an angle $phi$ with the $x$-axis, and let $phiin(0,2pi)$.
I want to argue that (maybe it is incorrect though)
$$Gamma(z)=int_{L}t^{z-1}e^{-t} dt$$
I firstly know that it converges when $Re(z)>0$. Choose a contour like sector, let $L_1={z=x+iy:y=0, r<x<R}$, $L_2={z=Re^{itheta}:0<theta<phi}$, $L_3={z=xe^{iphi}:r<x<R}$ and $L_4={z=re^{itheta}:0<theta<phi}$, where $r<R$ and the contour is counterclockwise. By Cauchy theorem we have the contour integral should be $0$.
Easy to see that (let $z=x+iy$)
$$
lim_{rto0+, Rto+infty}int_{L_1}t^{z-1}e^{-t} dt=Gamma(z)
$$
$$
|int_{L_4}t^{z-1}e^{-t} dt|=|int_{phi}^{0} e^{-re^{itheta}} (re^{itheta})^{z-1}ire^{itheta} dtheta| leq int_{0}^{phi} e^{-rcostheta} |r^{z}e^{itheta(z-1)}| dtheta=int_{0}^{phi} e^{-rcostheta} r^{x}e^{-theta y} dtheta to 0, r to 0+
$$
But when considering $L_2$:
$$
|int_{L_2}t^{z-1}e^{-t} dt|leqint_{0}^{phi} e^{-Rcostheta} R^{x}e^{-theta y} dtheta
$$
when for example, $frac{3pi}{2}>phi>frac{pi}{2}$, we have $costheta<0$ and I failed to prove the above integral goes to zero when $Rto +infty$.
Is my usage of Gamma function to compute the original integral a coincidence to get the correct result, or there is a way to prove my argument?
Thank you so much!
complex-analysis gamma-function
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
$endgroup$
1
$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07
add a comment |
$begingroup$
I want to compute
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx$$
using Gamma function.
I know that by change of variable, $y=sqrt{x}$, one gets
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx=2int_{0}^{+infty}sin y^2 dy=frac{sqrt{2pi}}{2}$$
by Fresnel's integral.
I try it by considering this:
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx$$
It converges for both real and imaginary part using Dirichlet test, and $0$ is not a problem here. Let the square root take the pricipal branch where $sqrt{1}=1$. Let $y=-ix$, then
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx=sqrt{i}int_{0}^{-iinfty}y^{-frac{1}{2}}e^{-y} dy=(frac{sqrt{2}}{2} + frac{sqrt{2}}{2}i)Gamma(frac{1}{2})=frac{sqrt{2pi}}{2} + frac{sqrt{2pi}}{2}i$$
And it coincides with the final answer!
My problem is, suppose $L$ is a ray starting from $0$ and has an angle $phi$ with the $x$-axis, and let $phiin(0,2pi)$.
I want to argue that (maybe it is incorrect though)
$$Gamma(z)=int_{L}t^{z-1}e^{-t} dt$$
I firstly know that it converges when $Re(z)>0$. Choose a contour like sector, let $L_1={z=x+iy:y=0, r<x<R}$, $L_2={z=Re^{itheta}:0<theta<phi}$, $L_3={z=xe^{iphi}:r<x<R}$ and $L_4={z=re^{itheta}:0<theta<phi}$, where $r<R$ and the contour is counterclockwise. By Cauchy theorem we have the contour integral should be $0$.
Easy to see that (let $z=x+iy$)
$$
lim_{rto0+, Rto+infty}int_{L_1}t^{z-1}e^{-t} dt=Gamma(z)
$$
$$
|int_{L_4}t^{z-1}e^{-t} dt|=|int_{phi}^{0} e^{-re^{itheta}} (re^{itheta})^{z-1}ire^{itheta} dtheta| leq int_{0}^{phi} e^{-rcostheta} |r^{z}e^{itheta(z-1)}| dtheta=int_{0}^{phi} e^{-rcostheta} r^{x}e^{-theta y} dtheta to 0, r to 0+
$$
But when considering $L_2$:
$$
|int_{L_2}t^{z-1}e^{-t} dt|leqint_{0}^{phi} e^{-Rcostheta} R^{x}e^{-theta y} dtheta
$$
when for example, $frac{3pi}{2}>phi>frac{pi}{2}$, we have $costheta<0$ and I failed to prove the above integral goes to zero when $Rto +infty$.
Is my usage of Gamma function to compute the original integral a coincidence to get the correct result, or there is a way to prove my argument?
Thank you so much!
complex-analysis gamma-function
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
$endgroup$
I want to compute
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx$$
using Gamma function.
I know that by change of variable, $y=sqrt{x}$, one gets
$$int_{0}^{+infty} frac{sin x}{sqrt{x}} dx=2int_{0}^{+infty}sin y^2 dy=frac{sqrt{2pi}}{2}$$
by Fresnel's integral.
I try it by considering this:
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx$$
It converges for both real and imaginary part using Dirichlet test, and $0$ is not a problem here. Let the square root take the pricipal branch where $sqrt{1}=1$. Let $y=-ix$, then
$$int_{0}^{+infty}x^{-frac{1}{2}}e^{ix} dx=sqrt{i}int_{0}^{-iinfty}y^{-frac{1}{2}}e^{-y} dy=(frac{sqrt{2}}{2} + frac{sqrt{2}}{2}i)Gamma(frac{1}{2})=frac{sqrt{2pi}}{2} + frac{sqrt{2pi}}{2}i$$
And it coincides with the final answer!
My problem is, suppose $L$ is a ray starting from $0$ and has an angle $phi$ with the $x$-axis, and let $phiin(0,2pi)$.
I want to argue that (maybe it is incorrect though)
$$Gamma(z)=int_{L}t^{z-1}e^{-t} dt$$
I firstly know that it converges when $Re(z)>0$. Choose a contour like sector, let $L_1={z=x+iy:y=0, r<x<R}$, $L_2={z=Re^{itheta}:0<theta<phi}$, $L_3={z=xe^{iphi}:r<x<R}$ and $L_4={z=re^{itheta}:0<theta<phi}$, where $r<R$ and the contour is counterclockwise. By Cauchy theorem we have the contour integral should be $0$.
Easy to see that (let $z=x+iy$)
$$
lim_{rto0+, Rto+infty}int_{L_1}t^{z-1}e^{-t} dt=Gamma(z)
$$
$$
|int_{L_4}t^{z-1}e^{-t} dt|=|int_{phi}^{0} e^{-re^{itheta}} (re^{itheta})^{z-1}ire^{itheta} dtheta| leq int_{0}^{phi} e^{-rcostheta} |r^{z}e^{itheta(z-1)}| dtheta=int_{0}^{phi} e^{-rcostheta} r^{x}e^{-theta y} dtheta to 0, r to 0+
$$
But when considering $L_2$:
$$
|int_{L_2}t^{z-1}e^{-t} dt|leqint_{0}^{phi} e^{-Rcostheta} R^{x}e^{-theta y} dtheta
$$
when for example, $frac{3pi}{2}>phi>frac{pi}{2}$, we have $costheta<0$ and I failed to prove the above integral goes to zero when $Rto +infty$.
Is my usage of Gamma function to compute the original integral a coincidence to get the correct result, or there is a way to prove my argument?
Thank you so much!
complex-analysis gamma-function
complex-analysis gamma-function
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
edited Jan 1 at 8:13
Edward Wang
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
asked Jan 1 at 7:52
Edward WangEdward Wang
823512
823512
1
$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07
add a comment |
1
$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07
1
1
$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07
$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07
add a comment |
1 Answer
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oldest
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$begingroup$
There's a bit of a problem because square roots are ambiguous. While $int_0^infty x^{-1/2}exp -zxdx=sqrt{pi}z^{-1/2}$ is easily proven on $Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?
Funnily enough, there's a completely different approach using the gamma function: $$int_0^inftyfrac{sin x dx}{sqrt{x}}=frac{1}{sqrt{pi}}Imint_0^infty dxint_0^infty y^{-1/2}exp -x(y-i)dy=frac{1}{sqrt{pi}}Imint_0^infty dyfrac{y^{-1/2}}{y-i}\=frac{1}{sqrt{pi}}int_0^inftyfrac{y^{-1/2}dy}{y^2+1}=frac{1}{sqrt{pi}}int_0^{pi/2}tan^{-1/2}theta dtheta\=frac{1}{2sqrt{pi}}operatorname{B}left(frac{1}{4},,frac{3}{4}right)=frac{1}{2sqrt{pi}}Gammaleft(frac{1}{4}right)Gammaleft(frac{3}{4}right)=frac{sqrt{pi}}{2}cscfrac{pi}{4}=sqrt{frac{pi}{2}}.$$
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
$endgroup$
add a comment |
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$begingroup$
There's a bit of a problem because square roots are ambiguous. While $int_0^infty x^{-1/2}exp -zxdx=sqrt{pi}z^{-1/2}$ is easily proven on $Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?
Funnily enough, there's a completely different approach using the gamma function: $$int_0^inftyfrac{sin x dx}{sqrt{x}}=frac{1}{sqrt{pi}}Imint_0^infty dxint_0^infty y^{-1/2}exp -x(y-i)dy=frac{1}{sqrt{pi}}Imint_0^infty dyfrac{y^{-1/2}}{y-i}\=frac{1}{sqrt{pi}}int_0^inftyfrac{y^{-1/2}dy}{y^2+1}=frac{1}{sqrt{pi}}int_0^{pi/2}tan^{-1/2}theta dtheta\=frac{1}{2sqrt{pi}}operatorname{B}left(frac{1}{4},,frac{3}{4}right)=frac{1}{2sqrt{pi}}Gammaleft(frac{1}{4}right)Gammaleft(frac{3}{4}right)=frac{sqrt{pi}}{2}cscfrac{pi}{4}=sqrt{frac{pi}{2}}.$$
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
$endgroup$
add a comment |
$begingroup$
There's a bit of a problem because square roots are ambiguous. While $int_0^infty x^{-1/2}exp -zxdx=sqrt{pi}z^{-1/2}$ is easily proven on $Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?
Funnily enough, there's a completely different approach using the gamma function: $$int_0^inftyfrac{sin x dx}{sqrt{x}}=frac{1}{sqrt{pi}}Imint_0^infty dxint_0^infty y^{-1/2}exp -x(y-i)dy=frac{1}{sqrt{pi}}Imint_0^infty dyfrac{y^{-1/2}}{y-i}\=frac{1}{sqrt{pi}}int_0^inftyfrac{y^{-1/2}dy}{y^2+1}=frac{1}{sqrt{pi}}int_0^{pi/2}tan^{-1/2}theta dtheta\=frac{1}{2sqrt{pi}}operatorname{B}left(frac{1}{4},,frac{3}{4}right)=frac{1}{2sqrt{pi}}Gammaleft(frac{1}{4}right)Gammaleft(frac{3}{4}right)=frac{sqrt{pi}}{2}cscfrac{pi}{4}=sqrt{frac{pi}{2}}.$$
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
$endgroup$
add a comment |
$begingroup$
There's a bit of a problem because square roots are ambiguous. While $int_0^infty x^{-1/2}exp -zxdx=sqrt{pi}z^{-1/2}$ is easily proven on $Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?
Funnily enough, there's a completely different approach using the gamma function: $$int_0^inftyfrac{sin x dx}{sqrt{x}}=frac{1}{sqrt{pi}}Imint_0^infty dxint_0^infty y^{-1/2}exp -x(y-i)dy=frac{1}{sqrt{pi}}Imint_0^infty dyfrac{y^{-1/2}}{y-i}\=frac{1}{sqrt{pi}}int_0^inftyfrac{y^{-1/2}dy}{y^2+1}=frac{1}{sqrt{pi}}int_0^{pi/2}tan^{-1/2}theta dtheta\=frac{1}{2sqrt{pi}}operatorname{B}left(frac{1}{4},,frac{3}{4}right)=frac{1}{2sqrt{pi}}Gammaleft(frac{1}{4}right)Gammaleft(frac{3}{4}right)=frac{sqrt{pi}}{2}cscfrac{pi}{4}=sqrt{frac{pi}{2}}.$$
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
$endgroup$
There's a bit of a problem because square roots are ambiguous. While $int_0^infty x^{-1/2}exp -zxdx=sqrt{pi}z^{-1/2}$ is easily proven on $Bbb R^+$, if you try an analytic continuation (whether your proof approach is as detailed as you've attempted or otherwise) the intended theorem isn't even clear. Which kind of $z^{-1/2}$ should be used for the desired case $z=-i$?
Funnily enough, there's a completely different approach using the gamma function: $$int_0^inftyfrac{sin x dx}{sqrt{x}}=frac{1}{sqrt{pi}}Imint_0^infty dxint_0^infty y^{-1/2}exp -x(y-i)dy=frac{1}{sqrt{pi}}Imint_0^infty dyfrac{y^{-1/2}}{y-i}\=frac{1}{sqrt{pi}}int_0^inftyfrac{y^{-1/2}dy}{y^2+1}=frac{1}{sqrt{pi}}int_0^{pi/2}tan^{-1/2}theta dtheta\=frac{1}{2sqrt{pi}}operatorname{B}left(frac{1}{4},,frac{3}{4}right)=frac{1}{2sqrt{pi}}Gammaleft(frac{1}{4}right)Gammaleft(frac{3}{4}right)=frac{sqrt{pi}}{2}cscfrac{pi}{4}=sqrt{frac{pi}{2}}.$$
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
answered Jan 1 at 8:51
J.G.J.G.
23.4k22237
23.4k22237
add a comment |
add a comment |
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$begingroup$
$int_0^infty sin y^2dy=int_0^inftysin(y^2)dyneint_0^inftysin^2ydy$.
$endgroup$
– Kemono Chen
Jan 1 at 8:07