Integral of 1/((x)e^x) ?

  • Thread starter S.N.
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  • #1
S.N.
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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.
 

Answers and Replies

  • #2
hunt_mat
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Maybe that is the final answer to your question, it's an integral.

What was the ODE?
 
  • #3
Char. Limit
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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 [tex]Ei(x) = \int_0^x \frac{e^t}{t} dt[/tex]
 
  • #4
S.N.
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great, thank you
 
  • #5
Dick
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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 [tex]Ei(x) = \int_0^x \frac{e^t}{t} dt[/tex]

I think it would actually be an 'incomplete gamma function', since the e^t is in the denominator.
 
  • #6
Char. Limit
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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:

[tex]\int - \frac{e^u}{u} du[/tex]

And the solution follows.
 
  • #7
Dick
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Actually, if you write this as e^(-t)/t, and then substitute u=-t, du=-dt, you get this:

[tex]\int - \frac{e^u}{u} du[/tex]

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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