nice-calculus-prove-that-0-ln-1-2-x-2-1-pi-2-x-2-dx-ln-pi-pi-golen-ratio-

Question Number 128707 by mnjuly1970 last updated on 09/Jan/21

\dots nice calculus \dots

p r o v e t h a t ::

\int_{0}^{\infty} \frac{l n (1 + φ^{2} x^{2})}{1 + π^{2} x^{2}} d x = l n (\frac{π + φ}{π})

φ ::= g o l e n r a t i o \dots

Answered by Dwaipayan Shikari last updated on 09/Jan/21

I (a) = \int_{0}^{\infty} \frac{l o g (1 + a^{2} x^{2})}{1 + π^{2} x^{2}} d x \Rightarrow I^{'} (a) = \int_{0}^{\infty} \frac{2 {a x}^{2}}{(1 + a^{2} x^{2}) (1 + π^{2} x^{2})} d x

= \frac{2 a}{a^{2} - π^{2}} \int_{0}^{\infty} \frac{1}{1 + π^{2} x^{2}} - \frac{1}{1 + a^{2} x^{2}} d x

= \frac{2 a}{a^{2} - π^{2}} (\frac{1}{π} . \frac{π}{2} - \frac{1}{a} . \frac{π}{2}) = \frac{1}{π + a}

I (a) = l o g (π + a) + C

C = - l o g (π) w h e n a = 0

I (a) = l o g (1 + \frac{a}{π}) \Rightarrow I (φ) = l o g (1 + \frac{φ}{π})

Commented by mnjuly1970 last updated on 09/Jan/21

s h o r t s o l u t i o n b u t

v e r y n i c e . t h a n k y o u \dots

Commented by Dwaipayan Shikari last updated on 09/Jan/21

W i t h p l e a s u r e s i r!