Is There a Solution to This Tricky Integral? 1/(xe^x)

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The integral of 1/(xe^x) is challenging and does not simplify easily using integration by parts. The discussion suggests that the solution involves the Exponential Integral function, Ei(x), defined as the integral from 0 to x of e^(t)/t dt. There is also mention of the incomplete gamma function, which can represent the same integral. A substitution method is proposed to transform the integral into a more manageable form. Ultimately, both the Exponential Integral and incomplete gamma function provide valid representations for the solution.
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Homework Statement


Find the integral of

1/(xe^x)


Homework Equations



None really... integration by parts maybe: integral of udv = vu - integral of vdu

The Attempt at a Solution



I tried this by parts but didn't really get anywhere, it definitely doesn't simplify into anything useful. It's the solution to an ODE so maybe there's a type in my book, because I always get this as the final integral I have to compute.
 
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Maybe that is the final answer to your question, it's an integral.

What was the ODE?
 
If you want an analytic answer, you'll need to use the Exponential Integral function Ei(x), defined to be the integral from 0 to x of e^(t)/t dt.

EDIT: This can be written as Ei(x) = \int_0^x \frac{e^t}{t} dt
 
great, thank you
 
Char. Limit said:
If you want an analytic answer, you'll need to use the Exponential Integral function Ei(x), defined to be the integral from 0 to x of e^(t)/t dt.

EDIT: This can be written as Ei(x) = \int_0^x \frac{e^t}{t} dt

I think it would actually be an 'incomplete gamma function', since the e^t is in the denominator.
 
Dick said:
I think it would actually be an 'incomplete gamma function', since the e^t is in the denominator.

Actually, if you write this as e^(-t)/t, and then substitute u=-t, du=-dt, you get this:

\int - \frac{e^u}{u} du

And the solution follows.
 
Char. Limit said:
Actually, if you write this as e^(-t)/t, and then substitute u=-t, du=-dt, you get this:

\int - \frac{e^u}{u} du

And the solution follows.

True. The incomplete gamma of degree 0 is basically the same as the Ei. You can represent it either way. Sorry.
 

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