Ttdfjt

I'll modify this part since I want the proof to be here.

$$sum_n=1^infty frac1n^2=frac43sum_n=0^infty frac1(2n+1)^2=-frac43sum_n=0^infty int_0^1 x^2nln x dx=frac43int_0^1 fracln xx^2-1dx$$
$$int_0^1 fracln xx^2-1dxoversetxrightarrow frac1x=int_1^infty fracln xx^2-1dxRightarrow sum_n=1^infty frac1n^2=frac23 int_0^infty fracln x(x+1)(x-1)dx$$

$$=frac23 I(1,-1)=frac23 fracln^2 (1)-ln^2(-1)2(1-(-1))=frac23 fracpi^24=fracpi^26$$

Where we considered the following integral:
$$I(a,b)=int_0^infty fracln x(x+a)(x+b)dxoversetxrightarrow fracabx=int_0^infty fraclnleft(fracabxright)(x+a)(x+b)dx$$
Summing up the two integrals from above gives:
$$2I(a,b)=ln(ab)int_0^infty frac1(x+a)(x+b)dx=fracln(ab)a-blnleft(fracx+bx+aright)bigg|_0^infty $$
$$Rightarrow I(a,b)=fracln(ab)2fraclnleft(fracabright)a-b=fracln^2 a-ln^2 b2(a-b)$$

From here we know for sure that

$$int_0^infty fracln x(x+a)(x-1)dx=fracln^2 a+pi^22(a+1) $$

But let's just say that I was brave enough and took $b=-1$ to get:
$$I(a,-1)=int_0^infty fracln x(x+a)(x-1)dx=fracln^2a -ln^2 (-1)2(a-(-1))=fracln^2 a+pi^2 2(a+1)$$

We already know that this is true from the linked post, but let's ignore this, since the linked post uses the Basel problem to prove the result.

Can someone prove rigorously that we are allowed to plug in $b=-1$ in order to get the correct result?

Fixing $a >0$, we see that the integral
beginalign*
f(z) := int_0^infty fraclog x(x+a)(x+z) dx.
endalign* 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
$$
f(z) = fraclog^2 a - log^2 z2(a-z)
$$ 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
$$
F(x) = int_1^x frac log t(t+a)(t-1) dt,quad xge 0
$$ so that $F(infty) - F(0) $ is the desired integral $displaystyle I(a) = int_0^infty fraclog x(x+a)(x-1) dx$. By integration by parts,
beginalign*
f(z) =& left[F(x)frac x-1x+zright]^infty _0 - (1+z)int_0^infty fracF(x)(x+z)^2 dx\
=& left(F(infty) +fracF(0)z right)- (1+z)int_0^infty fracF(x)(x+z)^2 dx,quad zneq 0
endalign* The first term tends to $I(a)$ as $zto -1$. For $z=-1+it, t>0$, the second term can be estimated as
beginalign*
|1+z|left|int_0^infty fracF(x)(x+z)^2 dxright| le& t int_0^infty frac dx \
le& int_0^infty fract(x-1)^2 + t^2 |F(x)| dx\
&xrightarrowtto 0 pi |F(1)| = 0.
endalign* Thus
$$
I(a)= lim_tto 0^+ f(-1+it) =fraclog^2 a +pi^22(1+a) .
$$
Note that this argument only works for $z=-1$ since $displaystyle F(x) =int_c^x fraclog t(t+a)(t-c) dt$ is not well-defined (due to the singularity at $t=c$) for $c>0, cne 1$.

1. In fact, the equation $$int_0^inftyfracln xmathrmdx(x+a)(x+b)=frac(ln a)^2-(ln b)^22(a-b)$$ holds for all distinct complex $a$, $b$ not lying on the nonpositive real axis, as follows immediately from the contour integral representation $$int_0^inftyfracln xmathrmdx(x+a)(x+b)=-int_Hfracmathrmdx2pimathrmifrac(ln x)^22(x-a)(x-b)text.$$ 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 $$mathrmPint_0^inftyfracln xmathrmdx(x+a)(x-lvert brvert)text.$$ In this case, we may use the Sokhotski–Plemelj theorem to arrive at $$mathrmPint_0^inftyfracln xmathrmdx(x+a)(x-lvert brvert)=frac(ln a)^2-(ln lvert brvert )^2+pi^22(a+lvert brvert)text.$$



3. If we let $b=-1$, then the integral converges and is equal to its Cauchy principal value. Therefore $$int_0^inftyfracln xmathrmdx(x+a)(x-1)=frac(ln a)^2+pi^22(a+1)text.$$
   

It is rather amusing that the substitution actually works.

Here is an 'elementary' proof of its validity, which stays clear of any complex analysis, infinite expansion, or special function such as Basel.

Decompose the integrand as below,

$$ I =int_0^infty fracln x(x+c)(x-1)dx = frac1c+1 int_0^infty ln xleft[ fracc-1(x+c)(x+1) +frac2x^2-1 right] dx $$

Thus, the integral $I$ can be expressed as,

$$ I = fracc-1c+1I_1 + frac2c+1I_2 tag1$$

where,

$$ I_1 =int_0^infty fracln x(x+c)(x+1)dx $$

and

$$ I_2 =int_0^infty fracln xx^2-1dx $$

The integral $I_1$, valid for $c>0$, has already been evaluated in a couple of places, which is sketched below with the variable change $x=c/t$,

$$ I_1 =int_0^infty fracln c - ln t(t+c)(t+1)dt = int_0^infty fracln c(t+c)(t+1)dt - I_1$$

$$ I_1 = fracln c2 int_0^infty fracdt(t+c)(t+1) = fracln^2 c2(c-1)tag2$$

The second integral is more involved, evaluated below with some agile manipulation. Recognizing,

$$ I_2 = J(alpha=1)$$

where $J(alpha)$ is defined as

$$ J(alpha) =int_0^infty fracln (1-alpha^2 + alpha^2 x^2)2(x^2-1)dx $$

The computation of its derivative is straightforward,

$$ J'(alpha) =int_0^infty fracalpha dx1-alpha^2 + alpha^2 x^2 = frac1sqrt1-alpha^2 int_0^infty fracdu1+u^2 = fractan^-1(infty)sqrt1-alpha^2 = fracpi/2sqrt1-alpha^2$$

which allows the desired integral $I_2$ to be evaluated,

$$ I_2 = J(1) = int_0^1 J'(alpha) dalpha = fracpi2int_0^1 fracdalphasqrt1-alpha^2 =fracpi2 sin^-1(1) = fracpi^24 tag3$$

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

$$ I =int_0^infty fracln x(x+c)(x-1)dx = fracpi^2 + ln^2 c2(c+1)$$