Let f(x) be an even function such that ∫0∞ f(x)dx=π2, then the value of  the definite integral ∫0∞ fx−1xdx is equal to

Let I=∫0∞ fx−1xdx…………..(1) Put x=1t⇒dx=−1t2dtFrom⁡(1):I=∫∞0 f1t−t−dtt2  Use ∫ab f(x)dx=−∫ba f(x)dx⇒I=∫0∞ f1t−tdtt2⇒I=∫0∞ ft−1tdtt2 (∵f is even function )⇒I=∫0∞ fx−1xdxx2…………...(2)Adding equations (1) & (2) ⇒2I=∫0∞ fx−1xdx+∫0∞ fx−1xdxx2⇒2I=∫0∞ fx−1x1+1x2dx Put x−1x=u⇒1+1x2dx=du∴2I=∫−∞∞ f(u)du ∵ If f is even ∫−aa f(x)=2×∫0a f(x)dx=2∫0∞ f(x)dx=2π2⇒2I=π∴I=π2