$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$

Question

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.

Ron Gordon · Answer 1 · 2015-08-20T16:55:49.630

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

$$\int_0^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) = 2 \int_2^{\infty} du \frac{f(u)}{\sqrt{u^2-4}} $$

+1 for a slick solution - for clarity's sake, the top integral here is the difference between the two terms in the OP's equation (using $\log x-\log a=\log\frac xa$). — Steven Stadnicki, Aug 20 '15 at 16:26

score 2 · Answer 2 · answered Aug 20 '15 at 16:20

2

Try to substitute $y=\frac{a^2}{x}$. Under such, one has: $\frac{a}{x}+\frac{x}{a}=\frac{y}{a}+\frac{a}{y}$.

answered Aug 20 '15 at 16:20

Damian

86

score 1 · Accepted Answer · answered Aug 20 '15 at 16:29

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 $\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$

$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$

3 Answers3