# int_0^oo arctanx/(x(x^2+a^2))dx ?

Jun 23, 2018

$I = \frac{\pi}{2 {a}^{2}} \ln \left(1 + \left\mid a \right\mid\right)$

#### Explanation:

We want to evaluate

$I = {\int}_{0}^{\infty} \arctan \frac{x}{x \left({x}^{2} + {a}^{2}\right)} \mathrm{dx}$

This can be solved by differentiation under the integral sign

Consider

$I \left(b\right) = {\int}_{0}^{\infty} \arctan \frac{b x}{x \left({x}^{2} + {a}^{2}\right)} \mathrm{dx}$

Notice color(blue)(I(0)=0 and color(blue)(I(1)=I

Differentiate both sides w.r.t. b

$I ' \left(b\right) = {\int}_{0}^{\infty} \frac{1}{\left({x}^{2} + {a}^{2}\right) \left({x}^{2} {b}^{2} + 1\right)} \mathrm{dx}$

By partial fractions

$I ' \left(b\right) = \frac{1}{1 - {a}^{2} {b}^{2}} {\int}_{0}^{\infty} \frac{1}{{x}^{2} + {a}^{2}} - {b}^{2} / \left({x}^{2} {b}^{2} + 1\right) \mathrm{dx}$

We recognize these integrals as

$I ' \left(b\right) = \frac{1}{1 - {a}^{2} {b}^{2}} {\left[\arctan \frac{\frac{x}{a}}{a} - b \arctan \left(b x\right)\right]}_{0}^{\infty}$

Assuming color(blue)(a>0 and color(blue)(b>=0

(We could just as well assumed color(blue)(a<0 )

$I ' \left(b\right) = \frac{1}{1 - {a}^{2} {b}^{2}} \left(\frac{\frac{\pi}{2}}{a} - b \frac{\pi}{2}\right)$

By some algebraic manipulation

$I ' \left(b\right) = \frac{\pi}{2} \cdot \frac{1}{1 - {a}^{2} {b}^{2}} \left(\frac{1 - a b}{a}\right)$

$\textcolor{w h i t e}{I ' \left(b\right)} = \frac{\pi}{2 a} \frac{1 - a b}{1 - {a}^{2} {b}^{2}}$

$\textcolor{w h i t e}{I ' \left(b\right)} = \frac{\pi}{2 a} \frac{1}{1 + a b}$

Integrate both sides w.r.t b

$I \left(b\right) = \frac{\pi}{2 a} \int \frac{1}{1 + a b} \mathrm{db}$

$\textcolor{w h i t e}{I \left(b\right)} = \frac{\pi}{2 {a}^{2}} \ln \left(1 + a b\right) + C$

Evaluate the constant of integration (Remember color(blue)(I(0)=0)

$0 = \frac{\pi}{2 {a}^{2}} \ln \left(1 + a \cdot 0\right) + C \implies C = 0$

Thus

$I \left(b\right) = \frac{\pi}{2 {a}^{2}} \ln \left(1 + a b\right)$

Respecting negative values of $a$ as well

$I \left(b\right) = \frac{\pi}{2 {a}^{2}} \ln \left(1 + \left\mid a \right\mid b\right)$

In your case color(blue)(b=1

$I \left(1\right) = I = \frac{\pi}{2 {a}^{2}} \ln \left(1 + \left\mid a \right\mid\right)$