# Why do we need a branch cut for $\int_0^{\infty} \frac{x^{\frac{1}{2}}}{{(1 + x)^2}}dx$?

What is the significance of the $x^{\frac{1}{2}}$ in the numerator of this integral. I have read this kind of integral requires taking a branch cut. Why do we need a branch cut, what does it enable us to do? What happens if we don't take one?

$$\int_0^{\infty} \frac{x^{\frac{1}{2}}}{{(1 + x)^2}}dx$$

Also, the pole in this problem lies on the real axis so we can use the method of "a semicircle $C_R$ and joined at its endpoints by a line on the real axis $C_L$...So what method do we use for such an integral?

• You need your function to be meromorphic, else all the magical wonders of countours and residues fail.
– Pedro
Commented Nov 14, 2014 at 14:47
• Put the branch cut on the positive real axis, that is, the argument of $z$ lies between $0$ and $2\pi$. Since $z^{1/2}$ is multi-valued, the integrals along $[0,\infty]$ do not cancel when we integrate along a keyhole contour. In the end we get $$\left(1-e^{\frac{1}{2}i2\pi}\right)\int^\infty_0\frac{\sqrt{x}}{(1+x)^2}{\rm d}x=2\pi i\left(\frac{1}{2i}\right)$$ and you may conclude. Commented Nov 14, 2014 at 15:00
• I still don't get it. What do you mean by the integrals don't cancel? Commented Nov 14, 2014 at 15:15
• As opposed to the integrals cancelling if we did what? Why must we use a branch cut?..what happens if we don't use one? Commented Nov 14, 2014 at 16:09
• We are simply not allowed to "not use a branch cut" as the function is multivalued and discontinuous about the branch point $z=0$. Using a keyhole contour with the cut on the positive real axis enables us to dodge the branch point singularity at $0$ and exploit the multivaluedness of the function to give us a result for the integral. On the other hand, if we were to integrate $\frac{1}{(1+z)^2}$, using a keyhole contour is meaningless, as the integrals will "cancel" and there is no need to avoid the point at $z=0$. Commented Nov 14, 2014 at 16:37

The reason we need a branch cut is because we need your integral to be single valued. Let $\sqrt{x}\mapsto\sqrt{z}$ where $z = re^{i\theta}$. Now if we encircle the branch point $z = 0$ by $2\pi$ we get $$\sqrt{r}e^{(i\theta + 2\pi i)/2} = \sqrt{r}e^{i\theta/2}e^{i\pi} = -\sqrt{r}e^{i\theta/2}$$ when we started at $\sqrt{r}e^{i\theta/2}$; therefore, the square root of $z$ is multiple valued and we need to define a branch cut in order to keep it single valued. Here is a key hole contour.

Now the pole of order two at $z = -1$ is enclosed in our contour. Let $f(z) = \frac{\sqrt{z}}{(1 + z)^2}$. Then \begin{align} \int_0^{\infty}f(z)dz &= \int_{\gamma}f(z)dz + \int_{\Gamma}f(z)dz + \int_{\text{upper line}}f(z)dz + \int_{\text{lower line}}f(z)dz\\ &= \frac{1}{2}\int_0^{\infty}f(z)dz + \frac{1}{2}\int_{\infty}^0-f(z)dz \end{align} By the estimation lemma, the first integral goes to zero as $\gamma\to 0$ and the second integral goes to zero as $\Gamma\to\infty$. We now have that the final two integrals are $2\pi i\sum\text{Res}$. Thus, $$\int_0^{\infty}f(z)dz = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$$

In my opinion it is premature to discuss branch cuts in relation to your integral. At this stage you are still working with real numbers, and it is quite possible that you can solve this integral without having to use the complex plane.

Clearly the first step in solving the integral is to get rid of the inconvenient $\sqrt{x}$ term. This can be done most easily by changing variables from $x$ to $t = \sqrt{x}$. So let us set $x = t^2$.

Note that the poles in the denominator are now located at $t = +i$ and $t = -i$. Furthermore the integrand is an even function of $t$, so you can extend the integration interval to $(-\infty, \infty)$ and use contour integration.

But I don't think this is necessary. You can make another substitution: $t = \tan(s)$. This way you can rid yourself completely of the denominator, and you are left with an integral from $0$ to $\pi/2$ of a real function that is simply the product of a few cosines and sines.

• I would point out the OP tagged his post with complex analysis and complex integration. In complex analysis books, they will say evaluate this real integral on $\mathbb{C}$ or something along those lines. So I would disagree with it is premature to discuss branch cuts. Commented Nov 15, 2014 at 2:59
• Okay, I see your point. But I assume that you agree with me that branch cuts only become relevant when a function is multi-valued. And in this case one can make the function single-valued. Commented Nov 15, 2014 at 6:35