# Integral

#### The Lord

##### New member
Prove that

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

#### chisigma

##### Well-known member
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

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I will start by the integral :

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

Now use the following substitution : $$x^2= t$$

$$\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt$$

By 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 :

$$F'(a) =\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma\left(\frac{a+1}{2}\right)\Gamma \left(2-\frac{a+1}{2}\right)\left[\psi \left(\frac{a+1}{2}\right)-\psi \left(2-\frac{a+1}{2}\right) \right]$$

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 :

$$\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx=-\frac{\pi}{4}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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 ...