Integral Calculus Question 105
Question: $ \int\limits_0^{\infty }{( \frac{\pi }{1+{{\pi }^{2}}x^{2}}-\frac{1}{1+x^{2}} )\log xdx} $ is equal to
Options:
A) $ -\frac{\pi }{2}\ln \pi $
B) 0
C) $ \frac{\pi }{2}\ln 2 $
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \int\limits_0^{\infty }{( \frac{\pi }{1+{{\pi }^{2}}x^{2}}-\frac{1}{1+x^{2}} )\log xdx} $ $ =\int\limits_0^{\infty }{\frac{\log ( \frac{y}{\pi } )dy}{1+y^{2}}=\int\limits_0^{\infty }{\frac{\log x}{1+x^{2}}dx}} $ $ =-\int\limits_0^{\infty }{\frac{\log \pi }{1+y^{2}}dy=-\frac{\pi }{2}In\pi } $
BETA
