An integral related to the digamma function I was playing around with asymptotics and integral formulas of the digamma function and exponential integral when I stumbled upon this one: $$\psi(x)=\int_0^\infty\left(\frac{e^{-t}}t-\frac{e^{-xt}}{1-e^{-t}}\right)~\mathrm dt,\quad\Re(x)>0$$ Working with it a bit, I ended up with the following integral: > $$I(a)=\int_1^\infty\frac{\ln(u-1)}{u^a}~\mathrm du,\quad\Re(a)>1$$ And I was wondering how to evaluate this. By letting $u\mapsto u+1$, $$I(a)=\int_0^\infty\frac{\ln(u)}{(u+1)^a}~\mathrm du$$ By letting $u\mapsto e^u$, $$I(a)=\int_{-\infty}^\infty\frac{ue^u}{(e^u+1)^a}~\mathrm du$$ Perhaps we can apply IBP, but that looks to be messy and involves limits. Likewise, the bounds and integrand don't look like they are very inviteful of a series expansion. Preferably, I'd like to solve this integral without the use of the digamma function.

It is not clear what you meant with "not using $\psi(x)$".

For $\Re(x) > 0$ $$\int_0^\infty\left(\frac{e^{-t}}t-\frac{e^{-xt}}{1-e^{-t}}\right)dt = \int_0^\infty e^{-t}\left(\frac{1}t-\frac{1}{1-e^{-t}}\right)dt+ \int_0^\infty\left(\frac{e^{-t}-e^{-xt}}{1-e^{-t}}\right)dt$$ $$ = C+\sum_{n=0}^\infty \int_0^\infty (e^{-(n+1)t}-e^{-(n+x) t})dt=C+\sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+x}\right) = C+\gamma + \psi(x)$$ $C=- \gamma$ by looking at $x=1$

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$$\int_1^\infty \ln(u-1)u^{-x}du = \frac{D}{x-1}+\int_1^\infty \ln(u-1)(u^{-x}-\frac{u^{-2}}{x-1})du \\\= \frac{1}{x-1}(D+\int_1^\infty \frac{u^{1-x}-u^{-1}}{u-1} du)=\frac{1}{x-1}(D+\int_0^\infty \frac{e^{(1-x)t}-e^{-t}}{1-e^{-t}} dt )= \frac{\psi(x-1)+D}{x-1}$$