How to integrate $\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}$ using the residue theorem.

Question

He was doing this integral using the formula $$\int_{0}^{\infty} \frac{dx}{\sqrt{x}(x^{2}+1)}= \frac{2\pi i}{1-e^{-2\pi i\alpha}}(\sum(Res(\frac{F(z)}{z^{\alpha}};z_{k})))$$ where $F(z)=\frac{1}{(x^{2}+1)}$, $\alpha=\frac{1}{2}$ and $z_{k}$ is a pole of $F(z)$. The poles of $F (z)$ are $i$ and $-i$ which are of order $1$. But it has not given me the answer which is $\frac{\pi}{\sqrt{2}}$.

I think my problem is calculating the residues or operate for to give me the answer. Any hint is appreciated.

Answer 1

Hint. You will need to consider which branch you take for the complex square root. In particular, $$(-i)^{1/2}=e^{-\pi i/4}$$ gives you the wrong answer, but $$(-i)^{1/2}=e^{3\pi i/4}$$ gives you the right answer. If you look carefully at the formula you are using where it was first stated, hopefully you will find an indication of what branch of $z^\alpha$ is to be used.

Answer 2

Here are some relevant hints including basic complex number arithmetic. Put $$\sqrt{z} = \exp(1/2\log z)$$

where the logarithm has the branch cut on the positive real axis (argument from zero to $2\pi$.)

Use a keyhole contour with the slot of the key on the positive real axis. On the circular part of the contour we have $1/\sqrt{z}/z^2\in\Theta(R^{-5/2})$ and since $\lim_{R\to\infty} 2\pi R \times R^{-5/2} = 0$ the contribution from the circular arc is zero.

This leaves the contribution above and below the positive real axis. Above the real axis we get $$\int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx$$ with $\sqrt{x}$ being the real square root function. Below the real axis we have $$\int_\infty^0 \frac{1}{\exp(1/2\log x + 1/2\times 2\pi i)(x^2+1)} dx$$ which is $$- e^{-\pi i} \int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx$$ with the square root again being the real square root function.

Collecting the two contributions we have $$(1-e^{-\pi i}) \int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx \\= 2\pi i \left(\mathrm{Res}\left(\frac{1}{\sqrt{z} (z^2+1)}; z=i\right) +\mathrm{Res}\left(\frac{1}{\sqrt{z} (z^2+1)}; z=-i\right)\right).$$

This implies that $$\int_0^\infty \frac{1}{\sqrt{x} (x^2+1)} dx = \pi i \left(\frac{1}{\sqrt{i} (+2i)} + \frac{1}{\sqrt{-i} (-2i)}\right) \\ = \pi i \left(\frac{1}{e^{i\pi/4} (+2i)} + \frac{1}{e^{3i\pi/4} (-2i)}\right) = \pi i e^{-i\pi /4} \left(\frac{1}{2i} - \frac{1}{2i e^{i\pi/2}}\right) \\= \pi i e^{-i\pi /4} \left(- i\frac{1}{2} + \frac{1}{2}\right) = \frac{\pi}{\sqrt{2}} i e^{-i\pi /4} \left(- i\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) \\ = \frac{\pi}{\sqrt{2}} i e^{-i\pi /4} e^{-i\pi /4} = \frac{\pi}{\sqrt{2}} i e^{-i\pi /2} = \frac{\pi}{\sqrt{2}} i (-i) \\ = \frac{\pi}{\sqrt{2}}.$$

Addendum. If we want to be extra careful about it we also have to check the integral along the miniature circle of radius $\epsilon$ enclosing the origin, which is on the order of $2\pi\epsilon/\sqrt{\epsilon}= 2\pi\sqrt{\epsilon}\to 0$ as $\epsilon\to 0.$