4

I am trying to evaluate the integral $\int_0^\infty \frac {x^{1/2} dx}{x^2 + 1}$ using method of residues. I can solve this very easily without the $x^{1/2}$ on top, but I do not know what to do when the function on top is not a polynomial where the powers are integers. How do you compute this?

Eugene Zhang
  • 16,805
Jerry Crowers
  • 109
  • 1
    keyhole contour – tired Nov 22 '15 at 23:58
  • Use a keyhole contour as suggested and $$\sqrt{z} = \exp(1/2\log(z))$$ with the branch cut of the logarithm on the positive real axis.Compute the residues at the poles at $z=\pm i$ and observe that you get a multiple of the target integral just below the real axis. – Marko Riedel Nov 23 '15 at 01:35
  • 1
    Or substitute $x = t^2$ and notice that the resulting integrand is even. Exploiting this symmetry, we can write $$ \int_{0}^{\infty} \frac{x^{1/2}}{1+x^2} , dx = \int_{-\infty}^{\infty} \frac{t^2}{t^4 + 1} , dt. $$ Now you can use the method of residue by considering upper-semicircular contour. Geometrically, we are kind of folding the keyhole contour (for the original integral). – Sangchul Lee Mar 27 '16 at 07:37

2 Answers2

2

Set $x=e^t$, we have $$ \int_0^\infty \frac {x^{1/2}}{x^2 + 1}dx=\int_{-\infty}^\infty \frac {e^{3/2t} }{e^{2t} + 1}dt=\int_{-\infty}^\infty \frac {e^{3/2x} }{e^{2x} + 1}dx $$ We consider the analytic function $$ F(z)=\frac {e^{3/2z} }{e^{2z} + 1} $$ on the rectangle contour of $C\cup C'\cup \gamma_1\cup \gamma_2$, where $C$ is $x$-axis from $-R$ to $R$, $C'$ is horizontal line from $R+\pi i$ to $-R+\pi i$, $\gamma_1$ is vertical line from $R$ to $R+\pi i$ and $\gamma_2$ is vertical line from $-R+\pi i$ to $-R$. By Cauchy's residue theorem, we have $$ \int_{-R}^RF(x)dx+\int_{C'}F(z)dz+\int_{\gamma_1}F(z)dz+\int_{\gamma_2}F(z)dz=2\pi iRes\left(F,\frac{\pi i}{2}\right)\tag1 $$ Note that only pole in the contour is $z=\frac{\pi i}{2}$. Since $$ Res\left(F,\frac{\pi i}{2}\right)=\lim_{z\to \frac{\pi i}{2}}(z-\frac{\pi i}{2})F(z)=\lim_{z\to \frac{\pi i}{2}}\frac{e^{3/2z}}{(e^{2z}+1)'}=-\frac1{2}e^{\frac{3\pi i}{4}}\tag2 $$ We have $$ \int_{C'}F(z)dz=\int_R^{-R}\frac{e^{3/2(x+\pi i)}}{e^{2(x+\pi i)}+1}\:dx=-e^{3/2\pi i}\int_{-R}^{R}\frac{e^{3/2x}}{e^{2x}+1}\:dx $$ Along with $$ \left|\int_{\gamma_1}F(z)dz\right|=\left|\int_{0}^{\pi}\frac{e^{3/2(R+iy )}}{e^{2(R+iy)}+1}idy\right|\leqslant \int_{0}^{\pi}\frac{|e^{3/2(R+iy )}|}{|e^{2(R+iy)}|-1}dy\leqslant \frac{\pi e^{3/2 R}}{e^{2R}-1}\to0 $$ as $R\to\infty$. And $$ \left|\int_{\gamma_2}F(z)dz\right|=\left|-\int_{0}^{\pi}\frac{e^{3/2(-R+iy )}}{e^{2(-R+iy)}+1}idy\right|\leqslant \int_{0}^{\pi}\frac{|e^{3/2(-R+iy )}|}{|e^{2(-R+iy)}|-1}dy\leqslant \frac{\pi e^{-3/2 R}}{1-e^{-2R}}\to0 $$ as $R\to\infty$. So by $(1),(2)$ $$ \lim_{R\to\infty}(1-e^{3/2\pi i})\int_{-R}^{R}\frac{e^{3/2x}}{e^{2x}+1}\:dx=-\pi ie^{\frac{3\pi i}{4}} $$ i.e. $$ \int_{-\infty}^{\infty}\frac{e^{3/2x}}{e^{2x}+1}\:dx=-\frac{\pi ie^{\frac{3\pi i}{4}}}{1-e^{3/2\pi i}}=\frac{\pi i}{e^{\frac{3\pi i}{4}}-e^{-\frac{3\pi i}{4}}}=\frac{\pi}{2\sin {\frac{3\pi }{4}}}=\frac{\pi}{\sqrt{2}} $$

Eugene Zhang
  • 16,805
0

Let

$$ I= \int_0^\infty \frac {x^{1/2}\, \mathrm{d}x}{x^2 + 1}. $$

By the Schwinger parametrization we have

$$ I=\int_0^\infty \mathrm{d}t\, \exp(-t)\int_0^\infty \mathrm{d}x\, \sqrt{x}\, \exp\left(-tx^2\right).$$

The last integral can be calculated along the lines of this answer. Using this result, one gets

$$I=\int_0^\infty \mathrm{d}t\,\frac{ \Gamma \left(\frac{3}{4}\right)\exp(-t)}{2 t^{3/4}}=\frac{1}{2}\Gamma \left(\frac{3}{4}\right)\Gamma \left(\frac{1}{4}\right)=\frac{\pi}{\sqrt{2}}$$

Dinesh Shankar
  • 840
  • 7
  • 21