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I want to evaluate the integral $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x}}{1+x^2}\,dx$ using complex analysis methods. I know that I have to use a keyhole contour, but I don't know which function to integrate on the contour. For example if I had to integrate $\displaystyle \int_{0}^{\infty}\frac{\ln x}{1+x^2}\,dx$ then I would have chosen to integrate over a keyhole contour the function $\displaystyle f(z)=\frac{\ln^2 z}{1+z^2}\,dx$.

If i had known that function then the rest is routine, since if we declare the given integral as $I$ and the contour integral as $J$ , then there is a simple relationship of $I, \; J$. It holds that $J=2\pi i I $. Hence we have the result.

However, I did not understand well the keyhole contour. If someone could show me how the other parts disappear I would be glad.

Tolaso
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2 Answers2

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Sometimes it is possible to remove branch cuts by using some clever (real) substitution.

In this case, by setting $x=y^2$ we have: $$ I = \int_{0}^{+\infty}\frac{\sqrt{x}}{1+x^2}\,dx = \int_{\mathbb{R}}\frac{y^2}{1+y^4}\,dy$$ and the last integral can be computed thorugh the residue theorem by using a semicircle in the upper half-plane as a contour, leading to: $$ I = 2\pi i\sum_{\xi\in\{\exp(\pi i/4),\exp(3\pi i/4)\}}\operatorname{Res}\left(\frac{1}{y^{-2}+y^2},y=\xi\right)=\color{red}{\frac{\pi}{\sqrt{2}}}.$$

Jack D'Aurizio
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  • Well, I know this kind of method.. but I wanted to see how to integrate on the keyhole contour.. I think I got it know... – Tolaso Jan 02 '15 at 15:35
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Integrate $$ f(z) = \frac{\sqrt{z}}{1+z^2} $$ along the keyhole contour. Choose the "natural" branch of $\sqrt{z}$ (the branch cut along the positive real axis), i.e. $\sqrt{z} = re^{it/2}$, where $z = re^{it}$ with $0 < t < 2\pi$. That way the two integrals along the positive real axis won't cancel.

mrf
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  • Thank you.. But I'm still missing one factor. The result should $\displaystyle \frac{\pi}{\sqrt{2}}$. While I get that $\sqrt{2}$ , I'm missing that $\pi $ factor. – Tolaso Jan 02 '15 at 14:27
  • The integral over the large circle will tend to $0$ as $R\to\infty$. The integral along the "lower" part of the real axis will pick up a minus sign from the branch and another minus sign from the orientation. So twice the integral you are looking for is $2\pi i$ times the sum of the residues of $f$ at $\pm i$. Remember to use the correct branch when you compute the residues. – mrf Jan 02 '15 at 14:29
  • Then I have to take the real part of the contour? Because obviously I'm getting something wrong here. I added the residues and got $\sum=\frac{1}{\sqrt{2}}$. And that $i$ did not get away. – Tolaso Jan 02 '15 at 14:37
  • Sorry.. I forgot about the branches.. Oups... – Tolaso Jan 02 '15 at 14:39