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

1 Answer
Jun 23, 2018

I=pi/(2a^2)ln(1+abs(a))

Explanation:

We want to evaluate

I=int_0^ooarctan(x)/(x(x^2+a^2))dx

This can be solved by differentiation under the integral sign

Consider

I(b)=int_0^ooarctan(bx)/(x(x^2+a^2))dx

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

Differentiate both sides w.r.t. b

I'(b)=int_0^oo1/((x^2+a^2)(x^2b^2+1))dx

By partial fractions

I'(b)=1/(1-a^2b^2)int_0^oo1/(x^2+a^2)-b^2/(x^2b^2+1)dx

We recognize these integrals as

I'(b)=1/(1-a^2b^2)[arctan(x/a)/a-b arctan(bx)]_0^oo

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

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

I'(b)=1/(1-a^2b^2)((pi/2)/a-bpi/2)

By some algebraic manipulation

I'(b)=pi/2*1/(1-a^2b^2)((1-ab)/a)

color(white)(I'(b))=pi/(2a)(1-ab)/(1-a^2b^2)

color(white)(I'(b))=pi/(2a)1/(1+ab)

Integrate both sides w.r.t b

I(b)=pi/(2a)int1/(1+ab)db

color(white)(I(b))=pi/(2a^2)ln(1+ab)+C

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

0=pi/(2a^2)ln(1+a*0)+C=>C=0

Thus

I(b)=pi/(2a^2)ln(1+ab)

Respecting negative values of a as well

I(b)=pi/(2a^2)ln(1+abs(a)b)

In your case color(blue)(b=1

I(1)=I=pi/(2a^2)ln(1+abs(a))