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Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\ln x\cdot\ln\left(1+\frac1{2\cosh x}\right)dx=\int_0^\infty\ln x\cdot\ln\left(1+\frac1{e^{-x}+e^x}\right)dx$$

I tried to evaluate it with a CAS, and looked up in integral tables, but was not successful.

OlegK
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2 Answers2

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Define

$$ I(\alpha) = \int_{0}^{\infty} \log x \log(1 - e^{-\alpha x}) \, dx. $$

Integrating by parts, followed by the substitution $\alpha x \mapsto x$, we have

\begin{align*} I(\alpha) &= \alpha \int_{0}^{\infty} \frac{x - x\log x}{e^{\alpha x} - 1} \, dx \\ &= \frac{1}{\alpha} \int_{0}^{\infty} \frac{(1+\log \alpha) x - x \log x}{e^{x} - 1} \, dx\\ &= \frac{1}{\alpha} \left\{ (1+\log\alpha)\zeta(2) - \left.\frac{d \zeta(s)\Gamma(s)}{s}\right|_{s=2} \right\}\\ &= \frac{1}{\alpha} \left\{ (\gamma+\log\alpha)\zeta(2) - \zeta'(2) \right\}. \end{align*}

Then it follows that

\begin{align*} \int_{0}^{\infty} \log x \log \left( 1 + \frac{1}{2\cosh x} \right) \, dx &= \int_{0}^{\infty} \log x \log \left( \frac{1 - e^{-3x}}{1 - e^{-x}} \cdot \frac{1 - e^{-2x}}{1 - e^{-4x}} \right) \, dx \\ &= I(2) + I(3) - I(1) - I(4) \\ &= \frac{5}{12} \zeta'(2) - \frac{5}{72}\gamma\pi^{2} + \frac{1}{18}\pi^{2} \log (3). \end{align*}

Plugging some identities relating $\zeta'(2)$ and the Glaisher-Kinkelin constant $A$, this reduces to Vladimir's answer.

Addendum - Something you might want to know:

The following identity played the key role in this proof.

$$ \int_{0}^{\infty} \frac{x^{s-1}}{e^{x} - 1} \, dx = \Gamma(s)\zeta(s), $$

which holds for $\Re s > 1$.

Sangchul Lee
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    As an aside, $$|\zeta^{(n)}(2)|=\sum_{k=1}^\infty\frac{\ln^nk}{k^2}\simeq\int_1^\infty\frac{\ln^nx}{x^2}dx=n!$$ and generally $$|\zeta^{(n)}(a)|=\sum_{k=1}^\infty\frac{\ln^nk}{k^a}\simeq\int_1^\infty\frac{\ln^nx}{x^a}dx=\frac{n!}{(a-1)^{n+1}}$$ where the integral expression for the factorial function from the first approximation is obtained from Euler's formula for the $\Gamma$ function, $n!=\int_0^1(-\ln x)^ndx=\int_0^1\ln^n\frac1xdx$ through the $t=\frac1x$ substitution. – Lucian Nov 09 '13 at 22:43
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I played around a little and noticed quickly that $$\displaystyle 1+\frac{1}{2}sech(x)=\frac{e^{2x}+x+1}{e^{2x}+1}$$

By noting the factorization $$\displaystyle \frac{1-x^{3}}{1-x}\cdot \frac{1-x^{2}}{1-x^{4}}=\frac{x^{2}+x+1}{x^{2}+1}$$

By letting $\displaystyle x\to e^{-x}$, we can break these up somewhat and write $$\displaystyle \frac{e^{2x}+e^{x}+1}{e^{2x}+1}=\frac{1-e^{-3x}}{1-e^{-x}}\cdot \frac{1-e^{-2x}}{1-e^{-4x}}$$

So, we can write $$\displaystyle log(x)\left[log\left(\frac{1-e^{-3x}}{1-e^{-x}}\right)+log\left(\frac{1-e^{-2x}}{1-e^{-4x}}\right)\right]$$

$$\displaystyle log(x)\left[log(1-e^{-3x})-log(1-e^{-x})+log(1-e^{-2x})-log(1-e^{-4x})\right]dx$$

Now, is there some sort of general form for something like $$\displaystyle \int_{0}^{\infty}log(x)log(1-e^{-nx})dx$$?.

If we use the series for $log(1+u)$, we get $$\displaystyle \int_{0}^{\infty}log(x)\sum_{k=1}^{\infty}\frac{(-1)^{k}e^{-nkx}}{k}dx$$

Now, it is beginning to look like the Gamma derivative $$\displaystyle \int_{0}^{\infty}x^{a-1}log(x)e^{-x}dx=\Gamma'(a)$$

I have it, I just need to put it together. I managed to derive the formula I have above.

One can easily evaluate $\displaystyle \int_{0}^{\infty}\frac{tlog(t)}{e^{t}-1}dt$ by using the identity

$$\displaystyle \int_{0}^{\infty}\frac{t^{s-1}}{e^{t}-1}dt=\Gamma(s)\zeta(s)$$ and diffing. Of course, you knew that.

Doing so, leads to the above needed integral and we get $$\displaystyle \int_{0}^{\infty}\frac{t^{s-1}log(t)}{e^{t}-1}dt=\Gamma(s)\zeta'(s)+\zeta(s)\Gamma'(s)$$

Let $s = 2$:

$$\displaystyle \int_{0}^{\infty}\frac{tlog(t)}{e^{t}-1}dt=\zeta'(2)+\frac{\pi^{2}}{6}(1-\gamma)$$

Our integral is $$\displaystyle \int_{0}^{\infty}log(x)log(1-e^{-nx})dx$$

Let $\displaystyle t=nx, \;\ dx=\frac{1}{n}dt$

$$\displaystyle \frac{1}{n}\int_{0}^{\infty}log(t)log(1-e^{-t})dt-\underbrace{\frac{1}{n}log(n)\int_{0}^{\infty}log(1-e^{-t})dt}_{\frac{1}{n}log(n)\frac{\pi^{2}}{6}}$$

Now, use parts: $$\displaystyle u=log(1-e^{-t}), \;\ dv=log(t), \;\ du=\frac{1}{e^{t}-1}dt, \;\ v=tlog(t)-t$$

$$\displaystyle \underbrace{\frac{1}{n}\cdot (tlog(t)-t)log(1-e^{-t})|_{0}^{\infty}}_{\text{0}}-\frac{1}{n}\int_{0}^{\infty}\frac{tlog(t)-t}{e^{t}-1}dt-\frac{1}{n}log(n)\frac{\pi^{2}}{6}$$

$$\displaystyle \frac{-1}{n}\int_{0}^{\infty}\frac{tlog(t)}{e^{t}-1}dt+\underbrace{\frac{1}{n}\int_{0}^{\infty}\frac{t}{e^{t}-1}dt}_{\frac{\pi^{2}}{6n}}-\frac{1}{n}log(n)\frac{\pi^{2}}{6}$$

$$\displaystyle \frac{-1}{n}\left[\zeta'(2)+\frac{\pi^{2}}{6}(1-\gamma)\right]+\frac{\pi^{2}}{6n}-\frac{\pi^{2}}{6n}log(n)$$

$$\displaystyle =\frac{\pi^{2}(\gamma-log(n))-6\zeta'(2)}{6n}$$

If one prefers in terms of log(A), then use the identity: $$\displaystyle \zeta'(2)=\frac{\pi^{2}}{6}\gamma+\frac{\pi^{2}}{6}log(2\pi)-2\pi^{2}log(A)$$

Martin.s
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