int_0^oo arctanx/(x(x^2+a^2))dx ?
1 Answer
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
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
(We could just as well assumed
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
0=pi/(2a^2)ln(1+a*0)+C=>C=0
Thus
I(b)=pi/(2a^2)ln(1+ab)
Respecting negative values of
I(b)=pi/(2a^2)ln(1+abs(a)b)
In your case
I(1)=I=pi/(2a^2)ln(1+abs(a))