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How can I compute $$\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}\quad?$$

My attempt: Looking at the limits of the integration I see that we should induce some $\tan^{-1}(x)$ so if we put $\infty$ we would get something like $\frac{\pi}{2}$ . But I am not sure how to proceed . Partial fractions don't yield good integral.

Daniel W. Farlow
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Archis Welankar
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2 Answers2

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For any $a > 0$, consider the integral

$$\mathcal{I} = \int_0^\infty \frac{1}{1+x^a} \frac{dx}{1+x^2}$$

Change variable to $y = \frac1x$, we have

$$\mathcal{I} = \int_0^\infty \frac{1}{1+y^{-a}}\frac{1}{1+y^{-2}} \frac{dy}{y^2} = \int_0^\infty \frac{y^a}{1+y^a}\frac{dy}{1+y^2}$$

Summing the two expression of $\mathcal{I}$ and divide by $2$, we get

$$\mathcal{I} = \frac12 \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{4}$$ independent of value of $a$.

Random Notes

About the question how do I arrive this answer, the key is symmetry.

When one is facing a complicated looking integral, the first thing one should do is looking for symmetry hidden in the integral.

For the integral at hand, the integrand has similar form and the range of integration is invariant under a change of variable from $x$ to $\frac1x$. In this case, there are a few things one should check.

  1. Do different part of integral cancel each other?
  2. Can the original and transformed integral combined together to something simpler?
  3. Will changing variable to $y = x \pm \frac{\text{const}}{x}$ reduces the integrand to a more manageable form.

It turns out the second strategy works and the rest is mostly following your nose.

achille hui
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  • Just like mine earlier. ;-)) – Mark Viola Aug 07 '16 at 17:14
  • Cute. Were the clues from the limits all you needed to spot this change of variables? – Morgan Rogers Aug 07 '16 at 17:16
  • @morganrogers The success of the methodology requires more than the limits extending from $0$ to $\infty$. – Mark Viola Aug 07 '16 at 17:20
  • I'm wondering if there's any way to see that the integral given by the derivative $dI/da$ is identically zero. – Spine Feast Aug 07 '16 at 17:21
  • Clearly that is true, but when evaluating definite integrals, the procedure is (as here) 'spot the trick and follow through'. You had to notice that the substitution given was potentially useful and then mess around a bit; I was wondering how you spotted it in the first place. – Morgan Rogers Aug 07 '16 at 17:23
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    @MorganRogers it is not the limits but rather the symmetry of the integrand. if you see an integrand who has similar form under change of variable $x \to \frac1x$, There are several things one should try and see whether it simplify the expression 1) Can they be combined together and 2) changing variable to $y = x \pm \frac{\text{const}}{x}$. – achille hui Aug 07 '16 at 17:24
  • @morganrogers I was able to discern quickly that the "trick" would be successful because of experience with integrals of similar attributes. And now you are equipped with this tool you can employ the next time you encounter an integral for which this tool is suited. -Mark – Mark Viola Aug 07 '16 at 17:37
  • Thanks, but I wasn't asking for myself! It would be helpful to include in an answer how you worked out to do what you did. This means that people can refer to this answer in future to answer similar enough problems. – Morgan Rogers Aug 07 '16 at 17:39
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HINT:

Enforce the substitution $x\to 1/x$ and combine the results.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Enforcing the substitution $x\to 1/x$ reveals $$\begin{align}I&=\int_0^\infty \frac{1}{(1+x^{2015})(1+x^2)}\,dx\\\\&=\int_0^\infty \frac{x^{2015}}{(1+x^{2015})(1+x^2)}\,dx\tag 1\end{align}$$Adding the right-hand and left-hand sides of $(1)$ and dividing by $2$ yields $$I=\frac12\int_0^\infty \frac{1}{1+x^2}\,dx=\pi/4$$

Mark Viola
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    +1. This is similar to a 'popular one' in MSE: $\displaystyle{\int_{0}^{\pi/2}{\mathrm{d}x \over 1 + \tan^{,\mu}\left(, x, \right)}}$. – Felix Marin Aug 08 '16 at 00:33
  • @felixmarin Thank you. And yes, you're right. Evaluating the integral you mentioned is facilitated by enforcing the substitution $x\to \pi/2 -x$. Then combining reduces to a trivial integral. It is indeed similar in kind. +1 for your very useful comment. -Mark – Mark Viola Aug 08 '16 at 01:43
  • You're welcome. My pleasure. – Mark Viola Aug 08 '16 at 16:58