I need to compute $$ \int_0^1 \frac{x-1}{\ln x}\,dx. $$ Using the fact that $\frac{d(x^y)}{dy} = (\ln x)x^y$,
I can't get any clue.
Can someone give me a tip on how to better approach solving these kinds of problems?
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Hint: Use Feynman's integration under the integral sign where the parameter is :
$$ \mathcal I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}\mathrm dx. $$
Workaholic
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Thnks~~~~~ !!! Could you please give me some tips about how to solve these kinds of tricky definite integrals? – user197137 Jan 22 '15 at 14:07
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@user197137 Practice. – Workaholic Jan 22 '15 at 14:33
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@user201168, remember what i said last time. – abel Jan 22 '15 at 14:43
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@abel What did you say? – Workaholic Jan 22 '15 at 15:02
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@user201168, that let no one teach you math. – abel Jan 22 '15 at 15:05
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@abel Ah! I remember quiet well, it's just that I thought it was McB who wrote that. ;-) – Workaholic Jan 22 '15 at 15:09
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1@user201168, this is from someone who teaches math for a living! – abel Jan 22 '15 at 15:18
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@abel Thanks, I really appreciate that. If you have more advice then please send it to me via my email (see my profile). – Workaholic Jan 22 '15 at 15:30
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Hint. You are on the right track, you can write $$\frac{x-1}{\ln x}=\int_{0}^{1}x^y dy$$ and then reverse the integration order.
Obtaining $$\int_0^1\frac{x-1}{\ln x}dx= \int_0^1\frac{1}{y+1}dy=\ln2$$
P. Appell
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Thnks~~~~~ !!! Could you please give me some tips about how to solve these kinds of tricky definite integrals? – user197137 Jan 22 '15 at 14:07
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The method of Differentiation Under the Integral Sign consists in differentiating $I(a)$ with respect to $a$ and then integrate it back with respect to $x$
Spoiler:
$$\begin{align}I(a) = \int_{0}^{1} \frac{x^a -1 }{\ln x} dx &\Rightarrow I'(a) = \int_{0}^{1} \frac{x^a\ln x }{\ln x} dx = \int_{0}^{1} x^a dx \\&\Rightarrow I(a) = \frac{x^{a+1}}{a+1}\Bigg|_{0}^{1}\end{align}$$
and then take $a = 1$.
Aaron Maroja
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