calculate-0-arctan-3x-arctan-2x-x-dx-

Question Number 68240 by mathmax by abdo last updated on 07/Sep/19

c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (3 x) - a r c t a n (2 x)}{x} d x

Commented by mathmax by abdo last updated on 08/Sep/19

l e t I = \int_{0}^{\infty} \frac{a r c t a n (3 x) - a r c t a n (2 x)}{x} d x a n d

I (ξ) = \int_{0}^{ξ} \frac{a r c t a n (3 x) - a r c t a n (2 x)}{x} d x w e h a v e {l i m}_{ξ \to + \infty} I (ξ) = I

I (ξ) = \int_{0}^{ξ} \frac{a r c t a n (3 x)}{x} d x - \int_{0}^{ξ} \frac{a r c t a n (2 x)}{x} d x b u t

\int_{0}^{ξ} \frac{a r c t a n (3 x)}{x} d x =_{3 x = t} \int_{0}^{3 ξ} \frac{a r c t a n (t)}{\frac{t}{3}} \frac{d t}{3} = \int_{0}^{3 ξ} \frac{a r c t a n (t)}{t} d t

\int_{0}^{ξ} \frac{a r c t a n (2 x)}{x} d x =_{2 x = t} \int_{0}^{2 ξ} \frac{a r c t a n (t)}{\frac{t}{2}} \frac{d t}{2} = \int_{0}^{2 ξ} \frac{a r c t a n (t)}{t} d t \Rightarrow

I (ξ) = \int_{0}^{3 ξ} \frac{a r c t a n (t)}{t} d t - \int_{0}^{2 ξ} \frac{a r c t a n t}{t} d t = \int_{2 ξ}^{3 ξ} \frac{a r c t a n (t)}{t} d t

\exists c \in] 2 ξ, 3 ξ [/ I (ξ) = a r c t a n (c) \int_{2 ξ}^{3 ξ} \frac{d t}{t} = a r c t a n (c) l n (\frac{3 ξ}{2 ξ}) \Rightarrow

{l i m}_{ξ \to + \infty} I (ξ) = \frac{π}{2} l n (\frac{3}{2}) \Rightarrow I = \frac{π}{2} l n (\frac{3}{2}) .