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Through a rather convoluted set of steps, I have been able to derive

$$\int_0^\infty \Big(\frac{\arcsin(\sin(t))}{\cosh(t)}\Big)^2dt = \frac{1}{2}\big(\frac{\pi^2}{3} - \pi\log{2} \big)$$

$$\int_0^\infty \Big(\frac{\arcsin(\sin(t))}{\cosh(t)}\Big)^4dt = \big(\frac{\pi^2}{6}-1\big)\big(\frac{\pi^2}{4} - \pi\log{2} \big) $$

I'm looking to extend the sequence to even powers of the integrand. My current method cannot be continued because it relied on closed-form expressions for $$ \sum_{m=1}^\infty \frac{(-1)^m}{m^{2k-1} \sinh{(\pi m x)} } \ , \ k=0, 1, 2 \ \text{ and } x=1. $$ As far as I'm aware, closed-form solutions for the previous equation with $k=3,4,...$ are not available. My method did not involve contour integration, which is something I infrequently use, so I'm hoping that more sophisticated methods might produce further examples.

user321120
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  • For $x=1$ and $k=4$, the value of the series is $- \frac{13 \pi^7}{453,600}$. If $x=1$, you can use contour integration to evaluate the series for any positive even value of $k$. – Random Variable Nov 06 '20 at 04:53
  • @RandomVariable Thanks for responding. The problem hasn't generated much interest, so if you write up your contour integration solution, say, for the integrand to the sixth power, I'll award 50 additional bonus points (although I've never done it before, and don't know how.) – user321120 Nov 06 '20 at 16:43

1 Answers1

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I assume the following approach is different from your approach since it's fairly straightforward.

Unfortunately it does not avoid having to evaluate alternating series involving the hyperbolic sine.

Taking the first integral as an example, we can expand $$s(x)= x^{2}, \ -\frac{\pi}{2} \le x \le \frac{\pi}{2} ,$$ in a Fourier series, replace $\left(\arcsin(\sin x)\right)^{2} $ with this series, switch the order of summation and integration, and then integrate.

Let's first find the coefficients for the Fourier series of $s(x)$.

By definition, $$a_{0} = \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} x^{2} \, \mathrm dx = \frac{\pi^{2}}{6},$$

$$a_{n}= \frac{2}{\pi} \int_{-\pi/2}^{\pi/2} x^{2} \cos(2nx) \, \mathrm dx = \frac{2}{\pi} \frac{\pi}{2n^{2}} \, \cos(\pi n)= \frac{(-1)^{n}}{n^{2}}, $$

and $$b_{n}= \frac{2}{\pi}\int_{-\pi/2}^{\pi/2} x^{2} \sin(2nx) \, \mathrm dx = 0.$$

Therefore, $$s(x) = \frac{\pi^{2}}{12} + \sum_{n=1}^{\infty} \frac{(-1)^{n}\cos(2nx)}{n^{2}}.$$

Replacing $\left(\arcsin(\sin x) \right)^{2} $ with this Fourier series and then changing the order of summation and integration, we get $$\int_{0}^{\infty} \frac{\left(\arcsin(\sin x) \right)^{2}}{\cosh^{2}(x)} \, \mathrm dx = \left(\frac{\pi^{2}}{12} \int_{0}^{\infty} \frac{\mathrm dx}{\cosh^{2}(x)} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \int_{0}^{\infty} \frac{\cos(2nx)}{\cosh^{2}(x)} \, \mathrm dx\right) . $$

The first integral is simple to evaluate since $\tanh(x)$ is an antiderivative of $\operatorname{sech}^{2}(x)$.

To evaluate the second integral, we can integrate the function $$f(z) = \frac{e^{2inx}}{\cosh^{2}(x)}$$ around a rectangular contour in the upper half of the complex plane of height $i \pi$.

We get $$\int_{-\infty}^{\infty} \frac{e^{2inx}}{\cosh^{2}(x)} \, \mathrm dx - e^{-2 n \pi}\int_{-\infty}^{\infty} \frac{e^{2inx}}{\cosh^{2}(x)} \, \mathrm dx = 2 \pi i \operatorname{Res} \left[f(z), \frac{i \pi}{2}\right] = 2 \pi i \left(2in e^{- \pi n} \right).$$

Then equating the real parts on both sides of the equation, we get $$\int_{-\infty}^{\infty}\frac{\cos(2nx)}{\cosh^{2}(x)} \, \mathrm dx = 2 \pi n \, \frac{2e^{- \pi n}}{1-e^{-2 \pi n}} = 2 \pi n \, \frac{2}{e^{\pi n}-e^{- \pi n}} = \frac{2 \pi n}{\sinh (\pi n)}. $$

Therefore, $$ \int_{0}^{\infty} \frac{\left(\arcsin(\sin x) \right)^{2}}{\cosh^{2}(x)} \, \mathrm dx = \frac{\pi^{2}}{12} + \pi \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n \sinh(\pi n)}.$$

I asked about that particular series (and related series) here.

Random Variable
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  • Yes, you got to the hyperbolic trig sum more efficiently than I did. I was hoping there might be a way to avoid those kind of sums altogether. I'll give it a few more days and then award the points and accept the answer, barring a breakthrough by another contributor. – user321120 Nov 07 '20 at 17:47
  • @skbmoore Even if no one posts another answer in the next few days, you don't need to award me a bounty or accept my answer. I'd be satisfied with a simple upvote if you found my answer in any way helpful or interesting. – Random Variable Nov 07 '20 at 18:13