The value of the integral _0^ _e(x)x^2+4dx is:

Question

The value of the integral $\int _{0}^{\infty}\frac{{\log }_{e}(x)}{{x}^{2}+4}dx$ is:

Accepted Answer

Let

$I={\int }_{\infty}^{0}\frac{\ln x}{{x}^{2}+4}\, dx$

We will use a standard substitution, as done in NCERT-type definite integral problems.

Take

$x=2t$

Then

$dx=2\, dt$

and when $x=0,t=0$, and when $x	o \infty,t	o \infty$.

So,

$I={\int }_{\infty}^{0}\frac{\ln (2t)}{(2t{)}^{2}+4}\cdot 2\, dt$

Simplify the denominator:

$(2t{)}^{2}+4=4{t}^{2}+4=4(1+{t}^{2})$

Hence,

$I={\int }_{\infty}^{0}\frac{\ln (2t)}{4(1+{t}^{2})}\cdot 2\, dt=\frac{1}{2}{\int }_{\infty}^{0}\frac{\ln (2t)}{1+{t}^{2}}\, dt$

Now use

$\ln (2t)=\ln 2+\ln t$

So,

$I=\frac{1}{2}{\int }_{\infty}^{0}\frac{\ln 2}{1+{t}^{2}}\, dt+\frac{1}{2}{\int }_{\infty}^{0}\frac{\ln t}{1+{t}^{2}}\, dt$

That is,

$I=\frac{\ln 2}{2}{\int }_{\infty}^{0}\frac{dt}{1+{t}^{2}}+\frac{1}{2}{\int }_{\infty}^{0}\frac{\ln t}{1+{t}^{2}}\, dt$

Now,

${\int }_{\infty}^{0}\frac{dt}{1+{t}^{2}}={[{	an }^{-1}t]}_{\infty}^{0}=\frac{\pi }{2}$

Also, let

$J={\int }_{\infty}^{0}\frac{\ln t}{1+{t}^{2}}\, dt$

Using the substitution $t=\frac{1}{u}$, we get

$dt=-\frac{1}{{u}^{2}}\, du$

Then

$J={\int }_{0}^{\infty}\frac{\ln (1/u)}{1+1/{u}^{2}}(-\frac{1}{{u}^{2}})du$

This simplifies to

$J={\int }_{\infty}^{0}\frac{-\ln u}{1+{u}^{2}}\, du=-J$

Therefore,

$J=0$

So,

$I=\frac{\ln 2}{2}\cdot \frac{\pi }{2}=\frac{\pi \ln 2}{4}$

Hence, the value of the integral is

$\frac{\pi \ln 2}{4}$

So, the correct option is:

Solutions · 1