# 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 $x=1$

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


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

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