# Homework Help: Improper integration change of variables

1. Jun 12, 2010

### Gregg

1. The problem statement, all variables and given/known data

show

$$\int _1^{\infty }\frac{1}{x^2}\text{Log}[x]dx=-\int_0^1 \text{Log}[x] \, dx$$

similarly show

$$\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx = 0$$

3. The attempt at a solution

For the first part a substitution 1/x works.

The second part I cannot do, I thought about

$$\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx$$

and then trying to maybe show

$$\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=-\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx$$

but for now I am not sure what to do.

2. Jun 12, 2010

### Dick

Try substituting u=1/x in one of your final two integrals.

3. Jun 12, 2010

### Gregg

$$\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^0\frac{1}{u^2+1}\text{Log}du$$

$$\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx=\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=0$$

thanks!