A lucky proof for the Basel problem.

Fixing $a >0$, we see that the integral \begin{align*} f(z) := \int_0^\infty \frac{\log x}{(x+a)(x+z)} dx. \end{align*} converges absolutely for $z$ away from $(-\infty,0]$. So $f$ defines an analytic function on $\mathbb C \setminus (-\infty,0]$ (which can be checked by Morera's theorem and Fubini's theorem, for example.) This implies $$ f(z) = \frac{\log^2 a - \log^2 z}{2(a-z)} $$ holds not only on $(0,\infty)$ but also $\mathbb C \setminus (-\infty,0]$ by analytic continuation. Now we need to check the continuity of $f$ at $z=-1$. Let $$ F(x) = \int_1^x \frac{ \log t}{(t+a)(t-1)} dt,\quad x\ge 0 $$ so that $F(\infty) - F(0) $ is the desired integral $\displaystyle I(a) = \int_0^\infty \frac{\log x}{(x+a)(x-1)} dx$. By integration by parts, \begin{align*} f(z) =& \left[F(x)\frac{ x-1}{x+z}\right]^\infty _0 - (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx\\ =& \left(F(\infty) +\frac{F(0)}{z} \right)- (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx,\quad z\neq 0 \end{align*} The first term tends to $I(a)$ as $z\to -1$. For $z=-1+it, t>0$, the second term can be estimated as \begin{align*} |1+z|\left|\int_0^\infty \frac{F(x)}{(x+z)^2} dx\right| \le& t \int_0^\infty \frac{|F(x)|}{|x-1+it|^2} dx \\ \le& \int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx\\ &\xrightarrow{t\to 0} \pi |F(1)| = 0. \end{align*} Thus $$ I(a)= \lim_{t\to 0^+} f(-1+it) =\frac{\log^2 a +\pi^2}{2(1+a)} . $$ Note that this argument only works for $z=-1$ since $\displaystyle F(x) =\int_c^x \frac{\log t}{(t+a)(t-c)} dt$ is not well-defined (due to the singularity at $t=c$) for $c>0, c\ne 1$.

1. In fact, the equation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=\frac{(\ln a)^2-(\ln b)^2}{2(a-b)}$$ holds for all distinct complex $a$, $b$ not lying on the nonpositive real axis, as follows immediately from the contour integral representation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=-\int_H\frac{\mathrm{d}x}{2\pi\mathrm{i}}\frac{(\ln x)^2}{2(x-a)(x-b)}\text{.}$$ Here $H$ is a Hankel contour about the negative real axis—where the branch cut of $\ln$ is chosen to be.

2. Even if $b<0$, we may consider the Cauchy principal value $$\mathrm{P}\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-\lvert b\rvert)}\text{.}$$ In this case, we may use the Sokhotski–Plemelj theorem to arrive at $$\mathrm{P}\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-\lvert b\rvert)}=\frac{(\ln a)^2-(\ln \lvert b\rvert )^2+\pi^2}{2(a+\lvert b\rvert)}\text{.}$$

3. If we let $b=-1$, then the integral converges and is equal to its Cauchy principal value. Therefore $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-1)}=\frac{(\ln a)^2+\pi^2}{2(a+1)}\text{.}$$

Here is an 'elementary' proof of its validity. Note $$ \frac1{(x+c)(x-1)}= \frac{1}{c+1} \left[ \frac{c-1}{(x+c)(x+1)} +\frac{2}{x^2-1} \right] $$

Then

$$ I= \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{c-1}{c+1}I_1 + \frac{2}{c+1}I_2 \tag{1}$$

with

$$ I_1 =\int_0^\infty \frac{\ln x \>dx }{(x+c)(x+1)} \overset{t=1/x}=\int_0^\infty \frac{\ln c \>dt}{(t+c)(t+1)}- I_1 = \frac{\ln^2 c}{2(c-1)}\tag{2}$$

and

$$ I_2 =\int_0^\infty \frac{\ln x}{x^2-1}dx = J(1)$$

where $J(\alpha)$ is defined as

$$ J(\alpha) =\int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{2(x^2-1)}dx $$

$$ J'(\alpha) =\int_0^\infty \frac{\alpha dx}{1-\alpha^2 + \alpha^2 x^2} = \frac{\pi/2}{\sqrt{1-\alpha^2}}$$

Then

$$ I_2 = J(1) = \int_0^1 J'(\alpha) d\alpha = \frac{\pi}{2}\int_0^1 \frac{d\alpha}{\sqrt{1-\alpha^2}} = \frac{\pi^2}{4} \tag{3}$$

Substitute (2) and (3) into (1), we obtain the sought-after result

$$ I =\int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{\pi^2 + \ln^2 c}{2(c+1)}$$