All Questions
6
questions
4
votes
3
answers
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I need help evaluating the integral $\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz$
I was playing around with the integral: $$\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz$$
I couldn't find a way of solving it, but I used WolframAlpha to find that the integral evaluated ...
0
votes
0
answers
50
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Further question on Logarithm product integral
How to perform $\int_0^1 \frac{\left(a_0\log(u)+a_1\log(1-u)+a_{2}\log(1-xu)\right)^9}{u-1} du $?
Method tried:
Intgration-by-parts
Series expansion
change of variable $\log(u)=x$
But I still can't ...
3
votes
1
answer
96
views
On $\int_0^{2\pi }\frac{\prod_{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod_{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx$
OP of this post evaluated a lot of remarkable polylog integrals (without proof), from which I conjecture a generalized one ($a_k\in \mathbb N$):
$$\int_0^{2 \pi } \frac{\prod _{k=1}^m \text{Li}_{a_k}(...
44
votes
2
answers
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Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$
We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
30
votes
4
answers
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Conjectural closed-form of $\int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$
Let $$I_n = \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$$
In a recently published article, $I_n$ are evaluated for $n\leq 6$:
$$\begin{aligned}I_1 &= \frac{\log ^2(2)}{2}-\frac{\pi ^2}{...
8
votes
0
answers
413
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More on the log sine integral $\int_0^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta$
I. In this post, the OP asks about the particular log sine integral,
$$\mathrm{Ls}_{7}^{\left ( 3 \right )} =-\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\...