0-ln-1-a-2-x-2-b-2-x-2-dx-

Question Number 151768 by mathdanisur last updated on 22/Aug/21

\underset{0}{\int^{\infty}} \frac{\ln (1 + a^{2} x^{2})}{b^{2} + x^{2}} dx = ?

Answered by Olaf_Thorendsen last updated on 22/Aug/21

f (a, b) = \int_{0}^{\infty} \frac{\ln (1 + a^{2} x^{2})}{b^{2} + x^{2}} d x (1)

\frac{\partial f}{\partial a} (a, b) = \int_{0}^{\infty} \frac{2 {a x}^{2}}{(1 + a^{2} x^{2}) (b^{2} + x^{2})} d x

\frac{\partial f}{\partial a} (a, b) = \frac{2 a}{a^{2} b^{2} - 1} \int_{0}^{\infty} (\frac{a^{2}}{1 + a^{2} x^{2}} - \frac{1}{b^{2} + x^{2}}) d x

\frac{\partial f}{\partial a} (a, b) = \frac{2 a}{a^{2} b^{2} - 1} {[a . \arctan (a x) - \frac{1}{b} \arctan (\frac{x}{b})]}_{0}^{\infty}

\frac{\partial f}{\partial a} (a, b) = \frac{2 a}{a^{2} b^{2} - 1} . \frac{π}{2} (a - \frac{1}{b})

\frac{\partial f}{\partial a} (a, b) = \frac{π (a^{2} - \frac{a}{b})}{a^{2} b^{2} - 1}

\frac{\partial f}{\partial a} (a, b) = \frac{π}{b^{2}} - \frac{π}{2 b^{3}} . \frac{2 {a b}^{2}}{a^{2} b^{2} - 1} + \frac{π}{b^{2}} . \frac{1}{a^{2} b^{2} - 1}

f (a, b) = \frac{π a}{b^{2}} - \frac{π}{2 b^{3}} \ln ∣ a^{2} b^{2} - 1 ∣ + \frac{π}{2 b^{3}} \ln ∣ \frac{a b - 1}{a b + 1} ∣ + C (b) (2)

with (1) : f (0, 1) = \int_{0}^{\infty} \frac{\ln (1 + 0 x^{2})}{1 + x^{2}} d x = 0

with (2) : f (0, 1) = C (b)

\Rightarrow C (b) = 0

f (a, b) = \frac{π a}{b^{2}} - \frac{π}{2 b^{3}} \ln ∣ a^{2} b^{2} - 1 ∣ + \frac{π}{2 b^{3}} \ln ∣ \frac{a b - 1}{a b + 1} ∣

f (a, b) = \frac{π a}{b^{2}} - \frac{π}{b^{3}} \ln ∣ a b + 1 ∣

Commented by mathdanisur last updated on 23/Aug/21

cool thank you Ser