What is this integral

Calculus Level 3

Evaluate

$\large \int_{0}^{\infty} \frac {(\ln x)^2 \ln {x^{2}}}{1+x^{2}} dx$

The answer is 0.

2 solutions

Chew-Seong Cheong
Apr 21, 2018

$\begin{aligned} I & = \int_0^\infty \frac {\ln^2 x \ln x^2}{1+x^2} dx \\ & = \int_0^\infty \frac {2\ln^3 x}{1+x^2} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left( \frac {2\ln^3 x}{1+x^2} + \frac {2\ln^3 \frac 1x}{x^2\left(1+\frac 1{x^2}\right)}\right) dx \\ & = \int_0^\infty \frac {\ln^3 x-\ln^3 x}{1+x^2} dx \\ & = \boxed{0} \end{aligned}$

Same way!!

Aaghaz Mahajan - 3 years, 1 month ago

How can I come up with the fomula in blue?

Tao Chen - 3 years ago

$\begin{aligned} I & = \int_0^\infty f(x) \ dx & \small \color{#3D99F6} \text{Let }u = \frac 1x \implies du = - \frac {dx}{x^2} \\ & = - \int_\infty^0 \frac {f\left(\frac 1u \right)}{u^2} du \\ & = \int^\infty_0 \frac {f\left(\frac 1u\right)}{u^2}du & \small \color{#3D99F6} \text{Replace }u \text{ with }x \\ & = \int^\infty_0 \frac {f\left(\frac 1x\right)}{x^2}dx \end{aligned}$

Also refer to "Inversion" in Integration Tricks .

Chew-Seong Cheong - 3 years ago

So, what kind of structure of the integral when I meet and I need to use this trick?

Tao Chen - 3 years ago

Robert Szafarczyk
Apr 20, 2018

Use a substitution $\frac {1}{u}=x$

What is this integral

The answer is 0.

2 solutions

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