For which $p>0$ does the following improper integral exist?

Question

The improper integral we have is $$\int^{1}_{0} |\ln{x}|^{p}dx$$ how do I approach this? I've never done anything like this and can't find any notes on it.

Thanks

Answer 1

Substitute $u=ln(x)$ to obtain,

$$\int_{-\infty}^0|u|^pe^udu$$

Then by integration by parts it follows,

$$\int_{-\infty}^0|u|^pe^udu=\int_{-\infty}^0(-u)^pe^udu=[(-u)^pe^u]_{-\infty}^0+p\int_{-\infty}^0(-u)^{p-1}e^udu$$ Repeat this $p$ times to get,

$$\int_{-\infty}^0|u|^pe^udu=p!$$

Answer 2

Let $$I_p=\int_{0}^{1}|\log x|^p\mathrm{d}x.$$

Note that $$I_p=(-1)^p\int_{0}^{1}(\log x)^p\;\mathrm{d}x.$$ Start by $p=1$:

$$I_1=\int_{0}^{1}|\log x|^1\mathrm{d}x=-\int_{0}^{1}\log x\;\mathrm{d}x=-\left[x\log x-x\right]_{0}^{1}=-[-1-0]=1.$$ For $p=2$:

$$I_2=\int_{0}^{1}|\log x|^2\mathrm{d}x=\int_{0}^{1}(\log x)^2\;\mathrm{d}x=\left[(\log x-1)x\log x\right]_{0}^{1}-\int_{0}^{1}(\log x-1)\;\mathrm{d}x=\dotsc$$

Is this helps you?