2

I'm trying to compute the integral $$\int_{0}^{\infty}dx \, x^{2}\operatorname{sech}^{2}(x)=\frac{\pi^{2}}{12}.$$ Manually, one obtains, quite naively, $$\int dx \, x^{2}\operatorname{sech}^{2}(x)=\operatorname{Li}_{2}(-e^{-2x})-x^{2}-2x\log(e^{-2x}+1)+x^{2}\tanh(x),$$ which of course is not possible to evaluate at infinity in this form.

I was trying to turn this thing into a series, but I'm having trouble.

ComplexYetTrivial
  • 12,094
KernelPanic
  • 45
  • Is this somehow related to you previous question, https://math.stackexchange.com/questions/3217348/evaluating-the-integral-x-ex1 where you assert that integral also has the value $\frac{\pi^2}{12}?$ – Thomas Andrews May 07 '19 at 16:44
  • @ThomasAndrews It is related but unrelated at the same time. They appear quite differently in my treatment of a problem but both happen to yield the same value. – KernelPanic May 07 '19 at 16:46
  • 1
    Note that it is possible to compute the limit of your indefinite integral as $x \to \infty$, since $\lim_{x \to \infty} x^2 (1 - \tanh (x)) = 0$ and $\lim_{x \to \infty} x \log(1 + \mathrm{e}^{-2x}) = 0$. Together with the lower limit this yields $- \operatorname{Li}_2 (-1) = \frac{\pi^2}{12}$ for the definite integral. – ComplexYetTrivial May 07 '19 at 20:13

2 Answers2

10

Since $\operatorname{sech}x=\frac{2e^{-x}}{1+e^{-2x}}$, your integral is $$\int_0^\infty\frac{4x^2e^{-2x}dx}{(1+e^{-2x})^2}=4\sum_{n\ge 1}(-1)^{n-1}n\int_0^\infty x^2e^{-2nx}dx=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n^2}=\eta(2)=\frac{\pi^2}{12}.$$

J.G.
  • 118,053
6

Sort of hopping off of @J.G.'s shoulders, we may evaluate $$f(s)=\int_0^\infty x^s\text{sech}^2(x) dx$$ using the same trick, namely $$f(s)=4\int_0^\infty \frac{x^s e^{-2x}dx}{(1+e^{-2x})^2}=4\sum_{n\geq1}(-1)^{n-1}n\int_0^\infty x^s e^{-2nx}dx$$ This integral is rather easy: $$\int_0^\infty x^se^{-2nx}dx=\frac{\Gamma(s+1)}{(2n)^{s+1}}$$ So $$f(s)=\frac{2^2}{2^{s+1}}\Gamma(s+1)\sum_{n\geq1}\frac{(-1)^{n-1}}{n^s}\\ f(s)= 2^{1-s}(1-2^{1-s})\Gamma(s+1)\zeta(s)$$

clathratus
  • 18,002