# Need a hint for this integral

I'm trying to evaluate the following integral

$$\int_0^{\infty} \frac{1}{x^{\frac{3}{2}}+1}\,dx.$$

This is an old complex analysis exam question, so I plan to use the residue theorem.

How can I first deal with the square-root, cubed term? I've been trying to find a clever substitution to reduce the problem to a simple one but so far I have not found a good one...

Any ideas are welcome.

Thanks,

• you could always use the substitution $u=\sqrt{x}$ and use partial fractions to get a couple rational functions with maximum degree of 2. Might not be the quickest solution, but it is what comes to mind – Brevan Ellefsen Jan 3 '16 at 23:10
• Hi @brevanellefsen - thanks for your comment, I will try this now... – User001 Jan 3 '16 at 23:11
• $du = -1/\sqrt{x}dx = -1/u dx$ so yes you will be getting a rational function. – mathreadler Jan 3 '16 at 23:14
• @mathreadler what? for the sub $u=\sqrt{x}$ you get $du = \frac{dx}{2\sqrt{x}} = \frac{dx}{2u}$ – Brevan Ellefsen Jan 3 '16 at 23:16
• Oh the effort. Well good luck any way. – mathreadler Jan 3 '16 at 23:22

Hint: Every time I see integrand like this which integrate over $(0,\infty)$, I will transform it to a beta integral and follows my nose.

Spoiler 1

Let $\displaystyle\;y = x^{3/2}\;$ and $\displaystyle\;z = \frac{y}{1+y}\;$, we have

Spoiler 2

\begin{align} \int_0^\infty \frac{dx}{x^{3/2}+1}&= \int_0^\infty \frac{dy^{2/3}}{y+1} = \frac23 \int_0^\infty \frac{y^{2/3-1} dy}{y+1}\\ &= \frac23 \int_0^\infty \left(\frac{y}{1+y}\right)^{2/3-1}\left(\frac{1}{1+y}\right)^{1/3-1}\frac{dy}{(1+y)^2}\\ &= \frac23 \int_0^1 z^{2/3-1}(1-z)^{1/3-1}dz \\ &= \frac23\frac{\Gamma(2/3)\Gamma(1/3)}{\Gamma(2/3+1/3)} = \frac23 \frac{\pi}{\sin\frac{\pi}{3}}\\ & = \frac{4\pi}{3\sqrt{3}} \end{align}

Update (sorry, this part is too hard to setup as spoiler correctly)

If you really want to compute the integral using residue, you can

1. change variable to $y = x^{3/2}$.
2. pick the branch cut of $y^{-1/3}$ in the resulting integrand along the positive real axis.
3. set up a integral over the contour:
$$C := +\infty - \epsilon i\quad\to\quad -\epsilon-\epsilon i\quad \to \quad -\epsilon + \epsilon i \quad\to\quad +\infty + \epsilon i$$

If you fix the argument of $y^{-1/3}$ to be zero on the upper branch of $C$, you will have

$$\left(1 - e^{-\frac{2\pi i}{3}}\right)\int_0^\infty \frac{y^{-1/3}dy}{y+1} = \int_C \frac{y^{-1/3}dy}{y+1} =^{\color{blue}{[1]}} 2\pi i\mathop{\text{Res}}_{y = -1}\left(\frac{y^{-1/3}}{y+1}\right) = 2\pi i e^{-\frac{\pi i}{3}}$$

This will give you $$\int_0^\infty \frac{dx}{x^{3/2}+1} = \frac23 \int_0^\infty \frac{y^{-1/3} dy}{y+1} = \frac{4\pi i}{3}\left(\frac{e^{-\frac{\pi i}{3}}}{1 - e^{-\frac{2\pi i}{3}}}\right) = \frac{2\pi}{3\sin\frac{\pi}{3}} = \frac{4\pi}{3\sqrt{3}}$$

Notes

• $\color{blue}{[1]}$ Since the integrand $\frac{y^{-1/3}}{y+1}$ goes to $0$ faster than $\frac{1}{|y|}$ as $|y| \to \infty$, we can complete the contour $C$ by a circle of infinite radius and evaluate the integral over $C$ by taking residue at poles within the extended contour.
• Interesting approach! – Brevan Ellefsen Jan 4 '16 at 0:03
• @User001 I have added an residue based approach. – achille hui Jan 4 '16 at 0:50
• @User001 Mine is equivalent to a keyhole contour. I don't want to describe the circle around origin as it wastes too many keystrokes for nothing. For your contour, I don't think evaluate it along the imaginary axis is easier than your original integral. – achille hui Jan 4 '16 at 1:40
• @User001 what form of integrand you are using? if you uses the one I use, it won't cancel out. There is a factor $e^{-\frac{2\pi i}{3}}$ coming from the $y^{-1/3}$. – achille hui Jan 4 '16 at 3:15
• @User001, if you want to use your integrand, you should use a contour like $[0,R] \cup \{ R e^{i\theta} : \theta \in [0,\frac{2\pi}{3}] \} \cup \{ r e^{i\frac{2\pi}{3}} : r \in [0,R] \}$. – achille hui Jan 4 '16 at 3:33

$$\int\frac{dx}{x^{3/2} + 1}$$ $$=2\int\frac{u}{u^3 + 1}du$$ Depending on how you want to solve the problem, you could start working from here using residue theory or save some work and break down into partial fractions $$= \frac{2}{3} \int \frac{u}{u^2 - u + 1}du + \frac{2}{3} \int \frac{1}{u^2 - u + 1}du - \frac{2}{3}\int \frac{1}{u+1}du$$ The first should fall to a $u$-sub, the second by completing the square (giving an answer in terms of $\arctan$) and the last is trivial. Of course, if you have to use residue theory than this is somewhat redundant :)

Let $t=\sqrt x$. Then $dt=\frac{1}{2}x^{-\frac{1}{2}}dx$, so $dx=2\sqrt x\ dt=2t\ dt$.

We have

$$\int\frac{2t}{t^3+1}dt$$

This integral is not going to get very simple as you can see via wolfram. However we can integrate by parts and use arctan.

$u=\frac{1}{t^3+1}$ and $dv=2t\ dt$

So $du=\frac{-3t^2}{(t^3+1)^2}dt$ and $v=t^2$

We have

$$\frac{t^2}{t^3+1}+\int \frac{3t^4}{(t^3+1)^2}$$

Now at least the term outside the integral is $0$, so we can focus on the integral. You should be able to use partial fractions at this point, and like I said there will be an arctan. I've checked this far with wolfram, by the way, but I'll leave the rest to you.

• Hmm... it almost seems easier to just use partial fractions than to use residue theory here. I took the same approach, more or less. – Brevan Ellefsen Jan 3 '16 at 23:18
• Im sure you're right. Like I said, judging from Wolfram's indefinite integral, its not going to get any easier than partial fractions. – Elliot G Jan 3 '16 at 23:22
• Hi @brevanellefsen, the integrand becomes an even function (odd / odd) and we can extend the integral over the whole real line, divide by two, and integrate, over an upper semi circular contour, which should enclose two poles. I think ... not 100% sure, but will try now :-) – User001 Jan 3 '16 at 23:27