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I'm currently studying for a complex analysis prelim. exam in August, so I'm working through some of the exercises in Titchmarsh. One of the exercises has us evaluate the integrals $$\int_0^\infty\frac{1}{1+x^4}\,dx\quad\text{and}\quad\int_0^\infty\frac{x^2}{1+x^4}\,dx.$$After evaluating each of them, I found $$\int_0^\infty\frac{1}{1+x^4}\,dx=\int_0^\infty\frac{x^2}{1+x^4}\,dx=\frac{\pi}{2\sqrt{2}}.$$Pretty sure I had miscalculated, I went to Wolfram Alpha to verify my answers only to find I had done it correctly.

My question is why these two have the same value. Intuitively, I expected $\int\frac{x^2}{1+x^4}\,dx$ to be larger because on the interval $(1,\infty)$, $x^2>1$. The only explanation I can think of is that the $x^2$ makes the integrand much smaller in the interval $[0,1]$ than the original function, but I wouldn't have guessed it to be enough to make the values come out the same. Is there some other intuitive reason why these two integrals are the same?

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    $\begingroup$ Look at their graphs. It seems there's some interval centered at $0$ outside of which the second function is always slightly larger than the first, but inside that interval, the first function is always significantly larger than the second. I suppose those two phenomena cancel out. That's hardly a complete answer, but it suggests a path for further investigation. $\endgroup$
    – Jack M
    Jul 14, 2013 at 20:07

2 Answers 2

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You may use the change of variables $x\leftrightarrow x^{-1}$ to verify the equality without evaluation.

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  • $\begingroup$ I figured there was a substitution like that which otherwise would make the identity work, but it still seems surprising to me. This makes for a nice verification of the identity, but it doesn't seem to offer any intuition as to why it's true... $\endgroup$
    – Clayton
    Jul 14, 2013 at 18:40
  • $\begingroup$ @Clayton What kind of explanation would you like to have? Maybe an example? $\endgroup$
    – Start wearing purple
    Jul 14, 2013 at 18:43
  • $\begingroup$ @Clayton: I have the same doubt as O.L., and let me consider your question more seriously. $\endgroup$
    – 23rd
    Jul 14, 2013 at 18:46
  • $\begingroup$ @O.L.: An explanation as to why the $x^2$ in the numerator doesn't make the integral any larger. It seems unusual to me that the $x^2$ is able to 'balance' the integrand perfectly to get the same value. $\endgroup$
    – Clayton
    Jul 14, 2013 at 18:46
  • $\begingroup$ If you want intuition as to how they could possibly be the same: when $0<x<1$, note that the LHS integrand is between $\frac{1}{2}$ and $1$, while the RHS integrand is between 0 and $\frac{1}{2}$; and yes, the RHS is larger for positive $x$, but they are BOTH very, very close to 0 for large $x$. $\endgroup$
    – Nick Peterson
    Jul 14, 2013 at 18:48
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To mitigate what you found counter-intuitiveness, computing

$$\int_0^1 (1+x^4)^{-1} \, dx \approx 0.867$$ $$\int_1^\infty (1+x^4)^{-1} \, dx \approx 0.244$$

shows that the majority of the area is between $[0,1]$, making it much less surprising that decreasing the value on $[0,1]$ could compensate increasing over $[1,\infty]$.

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