How to calculate integral $\int_1^{\infty} x^{-\theta} \log x \ \mathrm dx$?

Question

While I was reading an advanced maths stat textbook. It directly uses the following integral result for some proofs

$$\int_1^{\infty} x^{-\theta} \log x \ \mathrm dx=\dfrac 1{(\theta-1)^2}$$

I'm curious how to prove this. I tried normal tricks I know like doing integration by parts and substitutions. But none of them seems to work. Is there any other tricky way to do this.

Answer 1

Integration by parts with the choice $$u = \log x, \quad du = \frac{1}{x} \, dx, \\ dv = x^{-\theta} \, dx, \quad v = \frac{x^{-\theta+1}}{-\theta+1}$$ yields $$\int_{x=1}^\infty x^{-\theta} \log x \, dx = \left[\frac{x^{-\theta+1}}{-\theta+1} \log x \right]_{x=1}^\infty - \int_{x=1}^\infty \frac{x^{-\theta}}{-\theta+1} \, dx = \frac{1}{\theta-1} \int_{x=1}^\infty x^{-\theta} \, dx.$$ The rest is straightforward.

Answer 2

Note,

$$I(\theta)=\int_1^\infty x^{-\theta} dx = \frac1{\theta-1}$$ Thus,

$$\int_1^{\infty} x^{-\theta} \log x \ \mathrm dx =-I’(\theta)= \dfrac 1{(\theta-1)^2}$$

Answer 3

One could solve this by turning the integral into the form of the gamma function with a few substitutions. First $u=\ln(x)$, we turn it into

$$\int_0^\infty ue^{-u(\theta - 1)}du$$

Then, note that

$$\Gamma(x) = (x-1)! = \int_0^\infty t^{x-1}e^{-t}dt$$

So now with one final substitution $t=u(\theta - 1)$, we have

$$\frac{1}{(\theta-1)^2}\int_0^\infty te^{-t}dt = \frac{1}{(\theta-1)^2} (1!) = \frac{1}{(\theta-1)^2}$$