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I'm trying to solve $$\int_0^\infty\ln(x)\cdot \exp(x)\cdot x^{-x}\;\mathrm{d}x,$$ but I do not know how. Can someone give me a hint?

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Hint: $x^{-x} = e^{-x \ln(x)}$. Try a change of variables $u = x - x \ln(x)$.

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It is not allowed to change our original variable to any other name in a definite integral? – Babak S. Jun 28 '12 at 18:43
Who's not allowing it? – Robert Israel Jun 28 '12 at 19:13
I am asking if I have an definite integral like above; can I change my variable to any other name? I don't mean about your solution Prof. I ask generally. :) – Babak S. Jun 28 '12 at 19:20
In general, change of variable in a definite integral goes like this: $$\int_a^b f(g(t))\ g'(t)\ dt = \int_{g(a)}^{g(b)} f(u)\ du$$ – Robert Israel Jun 28 '12 at 21:34

$$\ln(x) \exp(x) x^{-x} = \ln(x) \exp(x) e^{-x \ln(x)} = \ln(x) \exp(x(1-\ln(x)))$$ and set $x - x \ln(x) = t$ and simplify to get a nice answer!

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HINT: First, $x^{-x}=e^{-x\ln x}$. Now combine the exponentials, and ask yourself what the derivative of $x-x\ln x$ is.

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$4$ answers in $46$ seconds! – user17762 Jun 28 '12 at 18:35

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