<s>[INST] $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ 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.
 [/INST] 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 $\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$
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