Marco Cantarini and Jack D'Aurizio proved hard-looking integrals (see Marco and Jack) in my recent two posts.

This is our final hard-looking integral that yield a rational answer:


Can anyone provide us a prove of $(1)$?

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    Both of these coincide with the integral $\displaystyle \int_0^\infty \frac{u \sin^4 u}{\cos u+\cosh u}\,du$, since they're obtained from it by substituting $u=x \pi$ and $u=x e$ respectively. – Semiclassical May 28 '16 at 18:00

3 Answers3


Taking the cue from Sophie Agnesi and doing a similar manner as my previous answer in your post, then \begin{align} S_A&=\int_{0}^{\infty}\frac{x\sin^4 x}{\cosh x+\cos x}\ dx\\[10pt] &=2\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty}x\ e^{-kx}\sin^3 x\sin kx\ dx\\[10pt] &=\frac{1}{2}\sum_{k=1}^\infty(-1)^{k-1}\left[3\int_{0}^{\infty}x\ e^{-kx}\sin x\sin kx\ dx-\int_{0}^{\infty}x\ e^{-kx}\sin3 x\sin kx\ dx\right]\\[10pt] &=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[\int_{0}^{\infty}x\ e^{-kx}\cos(k-3)x\ dx-\int_{0}^{\infty}x\ e^{-kx}\cos(k+3)x\ dx\right]\\[10pt] &=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[ \frac{\cos\left(2\tan^{-1}\left(\frac{k-3}{k}\right)\right)}{k^{2}+(k-3)^2}-\frac{\cos\left(2\tan^{-1}\left(\frac{k+3}{k}\right)\right)}{k^{2}+(k+3)^2}\right]\\[10pt] &=\frac{3}{4}-\frac{1}{4}\left(-\frac{29}{225}\right)\\[10pt] &=\frac{176}{225} \end{align} and the claim follows. I leave the latter sum to you as a brainstorming.

Anastasiya-Romanova 秀
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We may notice that: $$ \frac{1}{\cos x+\cosh x}=\frac{2\,e^{ix}}{\left(1+e^{(i+1)x}\right)\left(1+e^{(i-1)x}\right)}=\frac{2\,e^x}{\left(e^x+e^{ix}\right)\left(e^x+e^{-ix}\right)}\tag{1}$$ or: $$ \frac{1}{\cos x+\cosh x} = \frac{e^{ix}}{i\sin(x)}\cdot\frac{1}{e^x+e^{ix}}-\frac{e^{-ix}}{i\sin x}\cdot\frac{1}{e^x+e^{-ix}}\tag{1bis} $$ so that: $$ \frac{x \sin^4(x)}{\cos x+\cosh x} = \text{Re}\left[\frac{\frac{3x}{4}-\frac{x}{4}e^{-2ix}}{e^x+e^{ix}}+\frac{\frac{3x}{4}-\frac{x}{4}e^{2ix}}{e^x+e^{-ix}}\right]\tag{1ter} $$ but: $$ \int_{0}^{+\infty}\left(\frac{3x}{4}e^{ix}-\frac{x}{4}e^{3ix}\right) e^{-n(1+i)x}\,dx = \frac{1}{4}\left(\frac{3}{(-i+n(1+i))^2}-\frac{1}{(-3i+n(1+i))^2}\right)\tag{2} $$ so the whole problem boils down to computing: $$ \sum_{n\geq 1}(-1)^{n-1}\left(\frac{3(2n-1)}{(n^2+(n-1)^2)^2}-\frac{6n-9}{(n^2+(n-3)^2)^2}\right)\tag{3}$$ by properly exploiting the reflection formulas for $\psi$ and $\psi'$.

I think Marco's the expert at this point: his previous answer computes my $(3)$, too, and proves your conjecture. Anyway, I have an interesting remark: due to the particular form of the integrand function $(1)$, this kind of integrals can be directly converted into (relatively) simple series by regarding them as the integral remainder term in the Abel-Plana formula.

Jack D'Aurizio
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After making substitution $\pi x\mapsto x$ for the 1st integral and $e x\mapsto x$ for the 2nd integral, one will get

\begin{equation} \int_{0}^{\infty}\frac{x\sin^4x}{\cos x+\cosh x}\ dx \end{equation}

Now use the following relations

\begin{equation} \frac{\sin x}{\cos x+\cosh x} = 2\sum_{n=1}^{\infty}\ (-1)^{n-1} e^{-nx} \sin nx \end{equation}


\begin{equation} \sin^3x=\frac{3\sin x - \sin 3x}{4} \end{equation}

Apply the method used by user @MarcoCantarini in his answer then you should be able to get the desired result.

Sophie Agnesi
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