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The problem is to show

$$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$

for $t \gt 0$.

I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)

$$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$

but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.

Thanks a lot!

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1  
You have in fact a more general formula, see math.stackexchange.com/questions/61828/frullani-proof-integrals. –  Davide Giraudo Jun 29 '12 at 11:09

1 Answer 1

up vote 6 down vote accepted

Let $$I(t) = \int_0^\infty \dfrac{\exp(-x) - \exp(-xt)}{x} \, dx$$ Then $$\dfrac{dI}{dt} = \int_0^{\infty} \exp(-xt) \, dx = \left. \dfrac{\exp(-xt)}{-t} \right \vert_0^\infty = \dfrac1t$$ Hence, $$I(t) = \ln(t) + c$$ But $$I(1) = \int_0^\infty \dfrac{\exp(-x) - \exp(-x)}{x} \, dx = \int_0^\infty 0 \, dx = 0 \implies c =0$$ Hence, $$I(t) = \ln(t)$$

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Well, I was wanting a hint, but that's okay. Thank you. –  ae0709 Jun 28 '12 at 23:23
    
@platinumtucan I am sorry. I usually just look for the problem and go about solving it. I didn't read that you just wanted a hint. –  user17762 Jun 28 '12 at 23:28

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