How to calculate $\int_0^1 x^x\,dx$ using series? I read from a book that $$\int_0^1 x^x\,dx = 1-\frac{1}{2^2}+\frac{1}{3^3}+\dots+(-1)^n\frac{1}{(n+1)^{n+1}}+\cdots$$ but I can't prove it. Thanks in advance.
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@Ｊ.M. duplication confirmed xD, I browsed through this site first but all I found was the second link in my question... thanks! – arax Jun 01 '13 at 14:42
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No problem; it just happened that I've seen the "sophomore's dream" asked a number of times on this site... – J. M. ain't a mathematician Jun 01 '13 at 14:44
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Just write $$x^x=e^{x\ln x}=\sum_{n=0}^{\infty}\frac{(x\ln x)^n}{n!}$$ and use that $$\int_0^1(x\ln x)^n dx=\frac{(-1)^n n!}{(n+1)^{n+1}}.$$
To show the last formula, make the change of variables $x=e^{-y}$ so that $$\int_0^1(x\ln x)^n dx=(-1)^{n}\int_0^{\infty}y^ne^{-(n+1)y}dy,$$ which is clearly expressible in terms of the Gamma function.
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Nice, quick and simple, +1! Though you may want to elaborate on why the last equation is true. – dreamer Jun 01 '13 at 14:38
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I see that you have already added the part about making the change of variables now ;). That was what I meant. – dreamer Jun 02 '13 at 16:51