prove-0-a-ln-1-ax-1-x-2-dx-1-2-ln-1-a-2-tan-1-a-a-gt-0-

Question Number 98677 by PRITHWISH SEN 2 last updated on 15/Jun/20

prove

\int_{0}^{a} \frac{\ln (1 + ax)}{1 + x^{2}} dx = \frac{1}{2} \ln (1 + a^{2}) \tan^{- 1} a, a > 0

Answered by maths mind last updated on 15/Jun/20

f (t) = \int_{0}^{a} \frac{l n (1 + t x)}{(1 + x^{2})}

f^{'} (t) = \int_{0}^{a} \frac{x d x}{(1 + t x) (1 + x^{2})}

= \int_{0}^{a} \frac{(x + t) (1 + t x) - t (1 + x^{2})}{(1 + t^{2}) (1 + t x) (1 + x^{2})}

= \int_{0}^{a} \frac{x + t}{(1 + t^{2}) (1 + x^{2})} d x - \int_{0}^{a} \frac{t d x}{(1 + t^{2}) (1 + t x)}

= \frac{l n (1 + a^{2})}{2 (1 + t^{2})} + \frac{t a r c t a n (a)}{t^{2} + 1} - \frac{l n (1 + t a)}{1 + t^{2}}

f (0) = 0

f (a) = \int_{0}^{a} \frac{l n (1 + t a)}{1 + t^{2}} d t

f (a) = \int_{0}^{a} f (t) d t \Rightarrow f (a) = \int (\frac{l n (1 + a^{2})}{2 (1 + t^{2})} + \frac{t}{t^{2} + 1} a r c t a n (a) - \frac{l n (1 + t a)}{1 + t^{2}}) d t

\Rightarrow 2 f (a) = \int_{0}^{a} (\frac{l n (1 + a^{2})}{2 (1 + t^{2})} + \frac{t}{t^{2} + 1} a r c t a n (a)) d t

= \frac{l n (1 + a^{2})}{2} a r c t a n (a) + \frac{a r c t a n (a)}{2} l n (1 + a^{2})

\Leftrightarrow f (a) = \frac{l n (1 + a^{2})}{2} \tan^{- 1} (a)

Commented by PRITHWISH SEN 2 last updated on 16/Jun/20

thank you sir .