calculate-interms-of-a-and-b-the-integral-0-arctan-bt-arctan-at-t-dt-with-a-and-b-gt-0-

Question Number 31097 by abdo imad last updated on 02/Mar/18

c a l c u l a t e i n t e r m s o f a a n d b t h e i n t e g r a l

\int_{0}^{\infty} \frac{a r c t a n (b t) - a r c t a n (a t)}{t} d t w i t h a a n d b > 0 .

Commented by abdo imad last updated on 05/Mar/18

l e t p u t I = \int_{0}^{\infty} \frac{a r c t a n (b t) - a r c t a n (a t)}{t} d t a n d f o r ξ > 0

I (ξ) = \int_{0}^{ξ} \frac{a r c t a n (b t) - a r c t a n (a t)}{t} d t w e h a v e

I = {l i m}_{ξ \to + \infty} I (ξ) b u t I (ξ) = \int_{0}^{ξ} \frac{a r c t a n (b t)}{t} d t - \int_{0}^{ξ} \frac{a r c t a n (a t)}{t} d t

b u t \int_{0}^{ξ} \frac{a r c t a n (b t)}{t} d t =_{b t = x} \int_{0}^{b ξ} \frac{a r c t a n x}{x} d x

\int_{0}^{ξ} \frac{a r t a n (a t)}{t} d t =_{a t = x} \int_{0}^{a ξ} \frac{a r c t a n x}{x} d x \Rightarrow

I (ξ) = \int_{0}^{b ξ} \frac{a r c t a n x}{x} d x - \int_{0}^{a ξ} \frac{a r c t a n x}{x} d x = \int_{a ξ}^{b ξ} \frac{a r c t a n x}{x} d x

\exists c \in] a ξ, b ξ [/ I (ξ) = a r c t a n ξ \int_{a ξ}^{b ξ} \frac{d x}{x} = l n (\frac{b}{a}) a r c t a n ξ \Rightarrow

l i m_{ξ \to + \infty} I (ξ) = \frac{π}{2} l n (\frac{b}{a}) \Rightarrow I = \frac{π}{2} (l n (b) - l n (a)) .