How do I integrate $\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$? How do I integrate the following integral:
$$\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \mathrm{d}t$$
Where $a$ is some parameter?
I know that the solution includes Lerch Transcendents and logs (which is what I'm trying to arrive at); however, I've tried integrating this function, but failed.
I've already tried using simplifying it using series, which yielded a bunch of integrals as follows:
$$-\sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{t^k} {t^2+a^2} \mathrm{d}t - \frac{1}{a^2} \sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{t^k} {t^2+\frac{1}{a^2}} \mathrm{d}t + \frac{1}{a^2} \sum_{k \geq 1} \frac{(-1)^k}{k} \int_0^1 \frac{\log(t)} {t^2+\frac{1}{a^2}} \mathrm{d}t$$
However, I don't know how I'd proceed forwards from here.
 A: \begin{align}
&\int_0^{\infty} \frac{\log(t+1)}{t^2+a^2} \ dt\\
=& \int_0^{\infty}\int_0^1 \frac{t}{(t^2+a^2)(1+yt)}dy \ dt\\
=& \int_0^1 \frac1{1+a^2y^2}\int_0^\infty\left(\frac{a^2y}{t^2+a^2}+ \frac{t}{t^2+a^2}-\frac{y}{1+yt}\right)dt\ dy\\
= & \ \frac\pi2 \int_0^1 \frac{ay}{1+a^2y^2}dy-\int_0^1 \frac{\ln (a)}{1+a^2y^2}\ dy- \int_0^1 \frac{\ln (y)}{1+a^2y^2}\ dy\\
=& \ \frac\pi{4a}\ln(1+a^2)-\frac1a\ln (a)\tan^{-1}(a) 
-\frac i{2a}(\mathrm{Li}_2(ia)-\mathrm{Li}_2(-ia))
\end{align}
A: $$I(b)=\int_0^{\infty} \frac{\log(bt+1)}{t^2+a^2} \,dt$$
$$I'(b)=\int_0^{\infty} \frac{t}{\left(t^2+a^2\right) (b t+1)}\,dt=\frac{\pi  a b-2 \log (a b)}{2 \left(a^2 b^2+1\right)}$$
$$J=\int \frac{\pi  a b-2 \log (a b)}{2 \left(a^2 b^2+1\right)}\,db=\frac 1{2a}\int \frac{\pi  x-2 \log (x)}{x^2+1}\,dx$$
$$J=\frac 1{2a}\Bigg[\frac{1}{2} \pi  \log \left(x^2+1\right)-2 \log (x) \tan
   ^{-1}(x)+i (\text{Li}_2(-i x)-\text{Li}_2(i x)) \Bigg]$$ Go back to $b$, use the bounds to obtain what @Quanto already answered.
A: Partial Answer
By a counter integral we have
$$\int_{-\infty}^{\infty}\frac{\ln(t+1)}{t^2+a^2}dt=\frac{\pi}{2a}\ln(a^2+1)+\frac{\pi i}{a}\arctan a\tag 1$$
After some play, I got
$$\int_{-\infty}^0\frac{\ln(t+1)}{t^2+a^2}dt=\int_0^1\frac{\ln(1-t)}{t^2+a^2}dt+\int_0^1\frac{\ln(1-t)}{1+a^2t^2}dt-\int_0^1\frac{\ln t}{1+a^2t^2}dt+\frac{\pi i}{a}\arctan a\tag 2$$
From $(1)$ and $(2)$ and writing the $3$-rd integral in $(2)$ by means of "Li",
$$\int_{0}^{\infty}\frac{\ln(t+1)}{t^2+a^2}dt=\frac{\pi}{2a}\ln(a^2+1)+\frac{i}{2a}(\text{Li}_2(ai)-\text{Li}_2(-ai))-\int_0^1\frac{\ln(1-t)}{t^2+a^2}dt-\int_0^1\frac{\ln(1-t)}{1+a^2t^2}dt.$$
The last two integrals can also be evaluated by Li-function. I found them. I will update my answer if I can find some simplifications.
A: let's generalize your integral
$$ \mathcal{I}(b) = \int_0^\infty \frac{\ln(1+b \, x)}{a^2+x^2} \, \mathrm{d}x \hspace{5mm} \cdots (1) $$
Using Feynmann's Trick we get (differentiating w.r.t b)
\begin{align*} 
\mathcal{I}'(b) &= \int_0^\infty \frac{x}{(1+b\,x) \, (x^2 + a^2)} \, \mathrm{d}x \\
&= \frac{1}{1+a^2 \, b^2} \int_0^\infty \left( \frac{-b}{1+bx} + \frac{x}{x^2+a^2} + \frac{a^2 b}{x^2 + a^2} \right) \, \mathrm{d}x \\
&= \frac{1}{1+a^2 \, b^2} \left( - \ln(1+b \, x) + \frac{1}{2} \, \ln(x^2+a^2) + ab \tan^{-1} \left(\frac{x}{a} \right) \right) \bigg{|}_0^\infty \\
&= \frac{1}{1+a^2 \, b^2} \left( \ln \left( \frac{\sqrt{x^2+a^2}}{1+b x} \right)\bigg{|}_0^\infty + \frac{a b \pi}{2} \right) \\
&= \frac{1}{1+a^2 \, b^2} \left( - \ln(a) + \frac{a b \pi}{2} \right)
\end{align*}
Now we have
$$ \mathcal{I}'(b) = \frac{1}{1+a^2 \, b^2} \left( - \ln(a) + \frac{a b \pi}{2} 
\right) \hspace{5mm} \cdots (2)$$
After integrating (2) w.r.t b we get
$$ \mathcal{I}(b) = \frac{- \ln(a)}{a} \, \tan^{-1}(a b) + \frac{\pi}{4a} \ln(1 + a^2 b^2) + \mathrm{C} \hspace{5mm} \cdots (3) $$
Notice in (1) $$ \mathcal{I}(0) = 0 $$
So putting $b = 0$ in (3) we get
$$ \mathcal{I}(0) = 0 = 0 + 0 + \mathrm{C} \implies \mathrm{C} = 0 $$
So from (3)
$$ \mathcal{I}(b) = \frac{- \ln(a)}{a} \, \tan^{-1}(a b) + \frac{\pi}{4a} \ln(1 + a^2 b^2) \hspace{5mm} \cdots (4) $$
Now putting $b = 1$ in (4) we get desired answer
$$ \boxed{ \int_0^\infty \frac{\ln(1+x)}{a^2 + x^2} \, \mathrm{d}x = \frac{- \ln(a)}{a} \, \tan^{-1}(a) + \frac{\pi}{4a} \ln(1 + a^2) }$$
