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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.

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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 $x=1$


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

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left.\int_{0}^{\infty}\pars{{\expo{-t} \over t} - {\expo{-xt} \over 1 - \expo{-t}}}\,\dd t\,\right\vert_{\ \Re\pars{x}\ >\ 0} \,\,\,\stackrel{\exp\pars{-t}\ \mapsto\ t}{=} \int_{1}^{0}\bracks{{t \over -\ln\pars{t}} - {t^{x} \over 1 - t}}\, \pars{-\,{\dd t \over t}} \\[5mm] = &\ \int_{0}^{1}\bracks{{1 \over 1 - t}\int_{0}^{1}t^{y}\,\dd y - {t^{x - 1} \over 1 - t}}\,\dd t = \int_{0}^{1}\bracks{\int_{0}^{1}{1 - t^{x - 1} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{y} \over 1 - t}\,\dd t}\,\dd y \\[5mm] = &\ \int_{0}^{1}\bracks{\Psi\pars{x} - \Psi\pars{y + 1}}\,\dd y = \Psi\pars{x} - \ln\pars{\Gamma\pars{2} \over \Gamma\pars{1}} = \bbx{\Psi\pars{x}} \end{align}

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Let $u\mapsto1/u$ in the definition of $I(a)$ to get

$$I(a)=\int_0^1\ln\left(\frac1u-1\right)u^{a-2}~\mathrm du$$

Now note that:

$$\ln\left(\frac1u-1\right)=\ln(1-u)-\ln(u)$$

Thus, by series expansion,

$$\begin{align}\int_0^1\ln(1-u)u^{a-2}~\mathrm du&=-\sum_{k=1}^\infty\frac1k\int_0^1u^{a+k-2}~\mathrm du\\&=-\sum_{k=1}^\infty\frac1{k(a+k-1)}\\&=-\frac{\gamma+\psi(a)}{a-1}\end{align}$$

And by integration by parts,

$$\int_0^1\ln(u)u^{a-2}~\mathrm du=-\frac1{(a-1)^2}$$

So the given integral is

$$I(a)=\frac1{(a-1)^2}-\frac{\gamma+\psi(a)}{a-1}$$

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