Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 3 down vote favorite

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?

share|improve this question

3 Answers

up vote 3 down vote

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

share|improve this answer
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
2  
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
up vote 2 down vote

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

share|improve this answer
up vote 2 down vote

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

share|improve this answer
$4$ answers in $46$ seconds! – user17762 Jun 28 '12 at 18:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.