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
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do you need to know if it is finite or do you want to compute the integral? – Thorben Apr 11 '14 at 12:57
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If you set $u = \ln x$, i.e. $x = e^u$ then $dx = e^u du$ and you get $\int_{-\infty}^0 |u|^p e^u du$. – fgp Apr 11 '14 at 12:58
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@Thorben as it says in the title, I want to know for which values of $p>0$ this integral exists. And FTP how does that help? – user2850514 Apr 11 '14 at 13:18
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First I thought I can construct an integrable function wich bounds $|ln(x)|$ on $[0,1]$ to conclude that the integral must be finite. But i realised that the existence of such a bound is not so obvious... – Thorben Apr 11 '14 at 13:24
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This is Euler's first integral expression for the $\Gamma$ function. – Lucian Apr 11 '14 at 13:55
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@Lucian does it exist for all $p>0$? – user2850514 Apr 11 '14 at 13:59
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It exists for all $p>-1$. – Lucian Apr 11 '14 at 14:28
2 Answers
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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,
$$\int_{-\infty}^0|u|^pe^udu=p!$$
Thorben
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I understand what you've done so this clearly exists for every $p>0$ Right? Ps. There is no need for a substitution to get the same result – user2850514 Apr 11 '14 at 13:47
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Yes it seems to hold for every $p$. I know, but in this case I prefer this way to compute the integral... It is just a matter of preferences :D – Thorben Apr 11 '14 at 14:02
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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?
zighalo
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