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An approach bypassing polylogarithms is as follows: \begin{align}\int_1^{1+\sqrt2}\log\frac{t+1}{t(t-1)}\frac{dt}t&\stackrel{ibp}=\int_1^{1+\sqrt2}\frac{t^2+2t-1}{t(t^2-1)}\log t\,dt\\&\stackrel{t=e^u …

We seek \begin{align}I&=\int_0^\infty\frac{e^{-st}\sin t}{t(e^{at}+1)}\,dt\\&=\sum_{n\ge0}(-1)^n\int_0^\infty\frac{e^{-(s+a(n+1))t}\sin t}t\,dt=-\sum_{n\ge1}(-1)^n\arctan\frac1{s+an}\end{align} which …

Claim. $J(x)=\tfrac12x^2+\ln|x|+c$ for $|x|>1$ with $c$ constant. Proof: Here, an explicit form for $I(x)$ is not needed but I can't use it to prove $c=0$. For $|x|>1$, the substitution $y=2\cos t$ fo …

evaluate integrals of the form below $$\int_a^b\text{elementary function}(x)\,dx=\text{constant non-trivially involving}\,W(\cdot)\tag1$$ The elementary function must be continuous in the interval of integration … The constant, and limits of integration should not be engineered to contain $W$; e.g. writing $1$ as $W(k)/W(k)$, and expressing the constant as $W(ke^k)$ for some elementary $k$ is not accepted. …

On MSE I asked whether each of $\int_0^\infty\frac{dx}{x^x},\int_0^\infty\frac{dx}{x^{x^{x^x}}},\int_0^\infty\frac{dx}{x^{x^{x^{x^{x^x}}}}},\cdots$ was less than $2$ and received answers on bounding t …