MHB Prove the Integral: $$\frac{-\pi}{4}$$

[The Lord]

Prove that

$$ \int_0^\infty \frac{\log x}{(1+x^2)^2}dx = \frac{-\pi}{4}$$

 

[chisigma]

Prove that

$$ \int_0^\infty \frac{\log x}{(1+x^2)^2}dx = \frac{-\pi}{4}$$

First step may be to split the integral in two parts...

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{(1+x^{2})^{2}}\ dx = \int_{0}^{1} \frac{\ln x}{(1+x^{2})^{2}}\ dx + \int_{1}^{\infty} \frac{\ln x}{(1+x^{2})^{2}}\ dx = \int_{0}^{1} \frac{\ln x}{(1+x^{2})^{2}}\ dx - \int_{0}^{1} \frac{x^{2}\ \ln x}{(1+x^{2})^{2}}\ dx$ (1)

The second step is to remember the series expansion which holds for |x|<1...

$\displaystyle \frac{1}{(1+x^{2})^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\ (n+1)\ x^{2n}$ (2)

The third step is the following result I found some year ago... $\displaystyle \int_{0}^{1} x^{n}\ \ln x\ dx = - \frac{1}{(n+1)^{2}}$

Combining all these result we obtain...

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{(1+x^{2})^{2}}\ dx = - \sum_{n=0}^{\infty} (-1)^{n}\ \{\frac{1}{2n+1} - \frac{n}{(2n+1)^{2}} + \frac{n+2}{(2n + 3)^{2}} \} = - \sum_{n=0}^{\infty} \frac{1}{2n+1} = - \frac{\pi}{4}$ (3)

An interesting question : it is possible to obtain the same result using complex path integration?... in my opinion the answer is yes... but if we adopt the exact definition of the complex logarithm...Kind regards

$\chi$ $\sigma$

 

[alyafey22]

I will start by the integral :

\int^{\infty}_{0} \frac{x^a}{(1+x^2)^2}\, dx

Now use the following substitution : x^2= t\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dtBy the beta function this is equivalent to : \frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt = \frac{1}{2}B\left(\frac{a+1}{2},2-\frac{a+1}{2}\right)=\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(2-\frac{a+1}{2}\right)

Now let the following : F(a) =\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(2-\frac{a+1}{2}\right)Differentiate with respect to a :F&#039;(a) =\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma\left(\frac{a+1}{2}\right)\Gamma \left(2-\frac{a+1}{2}\right)\left[\psi \left(\frac{a+1}{2}\right)-\psi \left(2-\frac{a+1}{2}\right) \right]Now put a =0 and use \psi(x+1)=\psi(x)+\frac{1}{x}

\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)\left[\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)\right]=-\frac{\pi}{4}putting x^2= t we have our result :\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx=-\frac{\pi}{4}

 

[alyafey22]

An interesting question : it is possible to obtain the same result using complex path integration?... in my opinion the answer is yes... but if we adopt the exact definition of the complex logarithm...Kind regards

$\chi$ $\sigma$

Of course , but it is pretty tedious ...