Monday Math 73

Find $\int_0^1\frac{\ln{x}}{x+1}\,dx$ .

Here, we use the geometric series: as $\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k$ for $|x|\lt1$ , we have $\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$ for $|x|\lt1$
$\begin{eqnarray}\int_0^1\frac{\ln{x}}{x+1}\,dx&=&\int_0^{1}\ln(x)\sum_{n=0}^{\infty}(-1)^nx^n\,dx\\&=&\sum_{n=0}^{\infty}(-1)^n\int_0^{1}x^n\ln{x}\,dx\end{eqnarray}$ .
Now, we can find via integration by parts or by a table of integrals (see #3 here) that
$\int{x^n}\ln{x}\,dx=\frac{x^{n+1}}{n+1}\left(\ln{x}-\frac{1}{n+1}\right)+C$ , so
$\int_0^{1}x^n\ln{x}\,dx=-\frac{1}{(n+1)^2}$ (as $\lim_{x\to0+}x^p\ln{x}=0$ for any p>0).
Thus
$\begin{eqnarray}\int_0^1\frac{\ln{x}}{x+1}\,dx&=&\sum_{n=0}^{\infty}(-1)^n\int_0^{1}x^n\ln{x}\,dx\\&=&\sum_{n=0}^{\infty}(-1)^n\cdot\frac{-1}{(n+1)^2}\\&=&\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{(n+1)^2}\\&=&-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\\&=&-\eta(2)\\&=&-\frac{\pi^2}{12}\end{eqnarray}$ .

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Tags: Geometric Series, Integration, Math, Monday Math

This entry was posted on May 25, 2009 at 12:04 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Monday Math 73”

Avertedd Says:
May 26, 2009 at 12:01 am | Reply
У автора очень приятный слог
Monday Math 75 « Twisted One 151’s Weblog Says:
June 8, 2009 at 12:24 am | Reply
[...] then, is ? Note first that . Thus: . Reversing order of integration, . Now, as , , and so Now, I previously showed that ; thus [...]
Monday Math 80 « Twisted One 151’s Weblog Says:
July 13, 2009 at 12:05 am | Reply
[...] and one without. [More] Solution 1 (polylogarithm): We integrate directly in y: We showed here a way to find that . So now, we need to find . Making the substitution u=-x, we get . This is a [...]

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Monday Math 73

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