Non-converging integral that should be 0 A question requires you to prove that $$\int_0^\infty f(\frac{x}{a}+\frac{a}{x})\frac{\ln{x}}{x}dx=\ln{a}\int_0^\infty f(\frac{x}{a}+\frac{a}{x})\frac{1}{x}dx...........(1)$$
I made the substitution x=at, and split the resulting logarithm $\ln{at}$ in the integral into two, and got the above left-hand integral to be $$\ln{a}\int_0^\infty f(t+\frac{1}{t})\frac{1}{t}dt+\int_0^\infty f( t+\frac{1}{t})\frac{\ln{t}}{t}dt.......... (2)$$
Now, when you substitute back the variables in the first term above, you get the valie on the right in equation 1. So the second term in integral 2 must be 0.
But I can't integrate that second term, and when I ran that term on WolframAlpha, it came out as 0 for $f(x)=sin(x)$, but it doesn't even converge for $f(x)=cot(x)$ or $log(x)$.
So I've messed up somewhere there. Can someone help me with where and how?
 A: If you set $g(t) := f\left(t+ \frac{1}{t}\right)\frac{\ln(t) }{t}$
you can split the second term in (2) as
$$
\int_0^1 g(t)\,dt+ \int_1^{\infty} g(t)\,dt
$$
and observe that $\int_0^1 g\left(t\right)\,dt =-\int_1^{\infty} g(t)\,dt $
using the substitution $x=\frac{1}{t}$. 
(Here, $\int_0^1g(t)\,dt=\int_{\infty}^1 g(\frac{1}{t})\,d(\frac{1}{t})=\int_{\infty}^1f(t+\frac{1}{t})t (-\ln{t}).\frac{-1}{t^2}\,dt=-\int_1^{\infty}g(t)\,dt$).
Thus, summing up the two integrals you obtain $0$.
I think in the last two attempts you made the integral is not convergent, hence the two terms above are both infinite and the proof is no more true.
You should add the convergence of the integral in your hypothesis.
A: Substitute $x=\frac{a^2}t$
\begin{align}
I=& \int_0^\infty f(\frac{x}{a}+\frac{a}{x})\frac{\ln{x}}{x}dx\\
= &\int_0^\infty f(\frac{t}{a}+\frac{a}{t})\frac{\ln{a^2}-\ln t}{t}dt\\
= &2\ln{a}\int_0^\infty f(\frac{t}{a}+\frac{a}{t})\frac{1}{t}dt-I\\
= &\ln{a}\int_0^\infty f(\frac{t}{a}+\frac{a}{t})\frac{1}{t}dt
\end{align}
