I try to prove the following statement: $$\int\limits_{]0,\infty[}\frac{\ln{x}}{x^2-1} d\lambda_1(x)=\frac{\pi^2}{4}$$

There is also a clue: $$ \frac{1}{(1+y)(1+x^2y)}=\frac{1}{x^2-1}\left(\frac{x^2}{1+yx^2} - \frac{1}{1+y}\right)$$

$\ $

I tried to compute the Integral by partial integration and I get:

$$\int\limits_{]0,\infty[}\frac{\ln{x}}{x^2-1} d\lambda_1(x) = [ \ln{x}(\frac{1}{2}(\ln{(1-x)}-\ln{(x+1)}))]_0^\infty - \int\limits_{]0,\infty[}\frac{\frac{1}{2}(\ln{(1-x)}-\ln{(x+1)}))}{x} d\lambda_1(x)$$

But I don't thinks this is easier to handle. I thought maybe I could change the logarithm into a series but $x\in ]0,\infty[$ and not $|x-1|<1$.

I can't see the link between the $\ln{x}$ and $$\left(\frac{x^2}{1+yx^2} - \frac{1}{1+y}\right)$$

Why are there 2 variables?

I know how to compute $$\iint\limits_{]0,\infty[} \frac{1}{(1+y)(1+x^2y)}$$


  • $\begingroup$ What is $d\lambda_1(x)$? Is it just integration with respect to the Lebesgue measure? $\endgroup$ Dec 2, 2015 at 12:44
  • $\begingroup$ It's not mentioned but I would say yes. $\endgroup$
    – user185346
    Dec 2, 2015 at 13:01
  • $\begingroup$ This can be viewed as the derivative of the beta function, following a splitting of the integration interval into $(0,1)$ and $(1,\infty),$ and a substitution of the form $x=\dfrac1t$ on the latter. $\endgroup$
    – Lucian
    Dec 2, 2015 at 13:04
  • $\begingroup$ i think to use the given hint one have to employ some clever itnegral representation of $\log(x)$. this also explains where the second variable comes from $\endgroup$
    – tired
    Dec 2, 2015 at 13:18
  • $\begingroup$ Possible duplicate of Improper Integral $\int_0^1\frac{\ln(x)}{x^2-1}\,dx$ $\endgroup$
    – user91500
    Dec 2, 2015 at 13:54

1 Answer 1


Hint that will get you going (I think): $$ \int_0^{+\infty}\frac{1}{(1+y)(1+x^2y)}\,dy=\frac{2\ln x}{x^2-1}. $$ Now write things as a double integral, and change the order of integration.


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