$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ Show that $\displaystyle \int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
What substitution should i make for this.Both sides are looking alike,how to transform one
into another.Putting $\displaystyle \frac{a}{x}+\frac{x}{a}=t$ will not work,i think.
 A: No, but split the integral up into 2 pieces.  Consider the difference between the LHS and the RHS:
$$\int_0^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} = \\  \int_0^a \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} + \int_a^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} $$
In the latter integral, sub $x=a^2/u$ and you will find that the integral is the negative of the former integral.  viz.
$$\int_a^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} = \int_0^a \frac{du}{u} f \left ( \frac{a}{u} + \frac{u}{a} \right ) \log{\frac{a}{u}}$$
The integral over the infinite interval is thus zero, and the sought-after equality is true.
ADDENDUM
Not for nothing that 
$$\int_0^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) = 2 \int_2^{\infty} du \frac{f(u)}{\sqrt{u^2-4}} $$
A: Try to substitute $y=\frac{a^2}{x}$.
Under such, one has: $\frac{a}{x}+\frac{x}{a}=\frac{y}{a}+\frac{a}{y}$.
A: Let $$\displaystyle I = \int_{0}^{\infty} f\left(\frac{a}{x}+\frac{x}{a}\right)\cdot \frac{\ln x}{x}dx\;,$$ Now Put $\displaystyle \frac{a^2}{x} = t\;,$ Then $\displaystyle x= \frac{a^2}{t}$ and $\displaystyle dx = -\frac{a^2}{t^2}dt$
and Changing Limit, We get
$$\displaystyle I = \int_{\infty}^{0}f\left(\frac{t}{a}+\frac{a}{t}\right)\cdot \frac{t}{a^2}\cdot \ln\left(\frac{a^2}{t}\right)\cdot -\frac{a^2}{t^2}dt = \int_{0}^{\infty}f\left(\frac{t}{a}+\frac{a}{t}\right)\cdot \frac{1}{t}[2\ln a-\ln t]dt$$
So $$\displaystyle I = 2\ln a\int_{0}^{\infty}f\left(\frac{a}{t}+\frac{t}{a}\right)\cdot\frac{1}{t}-I\Rightarrow I = \ln a\int_{0}^{\infty}f\left(\frac{a}{t}+\frac{t}{a}\right)\cdot\frac{1}{t}dt$$
So $$\displaystyle I = \ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\cdot\frac{1}{x}dx$$
Using the formula $\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$
