How to solve $ \int_0^{\infty} \frac{x\ln(x)}{(1+x^2)^2}\,dx$?

Question

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

As the properties of definite integrals don't help I attempted to integrate by parts so

$\frac{\ln(x)}{2} \int \frac{2x}{({1+x^2})^2}dx $ is the first part which is $\frac{-\ln(x)}{2(x^2+1)}$ the second part is therefore $\ln(x) - \frac{\ln(1 + x^2)}{2}$ thus I= $\frac{-\ln(x)}{2(x^2+1)}- (\frac{-\ln(x)}{2(x^2+1)}$

however, this doesn't seem to make any sense as $\ln(\infty)$ isn't finite.

where am I going wrong?

Another approach: $$\int_0^{\infty} \frac{x\ln(x)}{(1+x^2)^2}dx\overset{x^2=t}{=}\frac{1}{4}\int_0^\infty\frac{\ln t}{(1+t)^2}dt$$ Next, analyse the substitution $t=\frac{1}{x}$ to evaluate the integral. — Svyatoslav, Sep 30 '22 at 13:18
but the integral still has a ln(x+1) term, so how doesn't it become finite? — math and physics forever, Sep 30 '22 at 13:23

score 1 · Accepted Answer · answered Sep 30 '22 at 13:30

1

Hint:

$1+x^2$ calls for $x=\tan y$

$$I=\int_0^{\pi/2}\dfrac{\tan y\ln(\tan y)}{1+\tan^2y}=\dfrac12\int_0^{\pi/2}\sin2y\ln(\tan y)\ dy$$

Using Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$., $$I=\dfrac12\int_0^{\pi/2}\sin2y\ln(\cot y)\ dy$$

Now $\ln(\cot y)=\cdots=-\ln(\tan y)$

answered Sep 30 '22 at 13:30

lab bhattacharjee

274,582

thanks a lot!!! – math and physics forever Sep 30 '22 at 13:36

How to solve $ \int_0^{\infty} \frac{x\ln(x)}{(1+x^2)^2}\,dx$?

1 Answers1