$\int_0^\infty$ $\frac{1}{1+x}$$\frac{dx}{\sqrt{x}}$

Part (a) asks to compute the integral by means of the residue at x = -1. I have done this just now, and the answer is $\pi$.

Part (b) asks, "can you see a simpler way to do it? explain."

I tried ordinary calculus methods, making two substitutions: once for the square root, and once for $tan(\theta)$, and get 1 in the integrand, after the trigonometric functions $\sec^2(\theta)$ cancel each other out. But this integral diverges, obviously.

So, what is the simpler way to do it, if it's not by ordinary calculus? Would it perhaps be another complex analysis method that's not so apparent?


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    $\begingroup$ Hint: try $x=\tan^2 t$ $\endgroup$ – Michael Galuza Jul 17 '15 at 3:56

You used the correct substitutions, but you forgot to change the bounds of the integral.

Here is how your approach should have worked out:

$\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+x)\sqrt{x}}\,dx$ $= \displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^2)t}2t\,dt$ $= \displaystyle\int_{0}^{\infty}\dfrac{2}{(1+t^2)}\,dt$

$= \displaystyle\int_{0}^{\pi/2}\dfrac{2}{1+\tan^2\theta}\sec^2\theta\,d\theta$ $= \displaystyle\int_{0}^{\pi/2}2\,d\theta = \pi$

The 2nd substitution was $t = \tan\theta$. When $t = 0$ we have $\theta = 0$ and when $t \to \infty$, we have $\theta \to \dfrac{\pi}{2}$. Thus, the bounds for the new integral are $0 \le \theta \le \dfrac{\pi}{2}$ and not $0$ to $\infty$. So this integral does not diverge.

Also, if you remember that $\dfrac{d}{dt}\arctan t = \dfrac{1}{1+t^2}$, then you can integrate $\displaystyle\int_{0}^{\infty}\dfrac{2}{(1+t^2)}\,dt$ without the need for a substitution.

  • $\begingroup$ Ah, I was close, @JimmyK4542. Can I ask you a silly question regarding the upper bound being pi/2? Why can't I choose another (positive) value for theta, which will also make tan blow up to infinity? Since tan is $\pi$-periodic, why can't I choose pi/2 + pi as my upper limit? I would still have the right bounds...but now my integral will yield a different value... $\endgroup$ – User001 Jul 17 '15 at 4:36
  • $\begingroup$ Way back in Calc I, we were just taught that arctan is not defined outside of -pi/2 and pi/2, so we must restrict to this interval - is that the rule that forces my upper limit to be pi/2 and not anything bigger, @jimmyk4542? $\endgroup$ – User001 Jul 17 '15 at 4:39
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    $\begingroup$ @LebronJames : The Fundamental Theorem of Calculus (which is being used here) requires that the integrand be continuous on the (open) interval of integration. (For an example why, consider the step function, which has derivative zero everywhere but is not constant.) So the upper bound cannot be any further away than the first point where the integrand is discontinuous. Entirely separately, going to one of the other upper limits means you're summing up multiple copies of your integral and will get too large a result. (The circumference is not $2\pi r$ if you go around the circle twice...) $\endgroup$ – Eric Towers Jul 17 '15 at 4:51
  • $\begingroup$ Hi @EricTowers, thanks for this amazingly clear and awesome explanation! Have a great night :-) $\endgroup$ – User001 Jul 17 '15 at 5:07

Let $x = y^2$. We have: $$\int_0^\infty \frac{1}{1+x}\frac{1}{\sqrt{x}}\,{\rm d}x = \int_0^\infty \frac{2}{1+y^2}\,{\rm d}y = 2 \arctan y \big|_0^\infty = 2\frac{\pi}{2} = \pi.$$

  • 1
    $\begingroup$ Thanks so much for the quick reply, @IvoTerek!! :-) $\endgroup$ – User001 Jul 17 '15 at 4:37

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