nasty integral

v71 · September 7th 2006, 02:04 PM

which is the integral for exp( -a/x ) dx ?

i have tried every method

thanks

**topsquark** · September 7th 2006, 03:11 PM

Originally Posted by v71

which is the integral for exp( -a/x ) dx ?

i have tried every method

thanks

That's because the indefinite form is...ugly. I don't think it has a closed form. If this is supposed to be a definite integral, though, you may be able to calculate an exact value.

-Dan

**ThePerfectHacker** · September 7th 2006, 04:34 PM

Topsquark is correct.
---
It is a famous case that,
$\int \frac{e^x}{x}dx$ is non-elementary.
We shall convert this integral into this form.
For simplicity sake we shall use $a=1$ , thus,
$\int e^{-1/x}dx$
Manipulate it as,
$\int \frac{x^2e^{1/x}}{x^2}dx$
Let, (going to apply the substitution rule)
$u=-1/x$ then, $u'=1/x^2$
Thus,
$\int \frac{e^u}{u^2}du$
If you use integration by parts with,
$v=e^u$ and $w'=1/u^2$
Then,
$-\frac{e^u}{u}+\int \frac{e^u}{u}du$
We have converted this integral into an integral which has no elementary anti-derivative. This confirms topsquarks guess.

**topsquark** · September 7th 2006, 04:44 PM

Originally Posted by ThePerfectHacker

Topsquark is correct.
---
It is a famous case that,
$\int \frac{e^x}{x}dx$ is non-elementary.
We shall convert this integral into this form.
For simplicity sake we shall use $a=1$ , thus,
$\int e^{-1/x}dx$
Manipulate it as,
$\int \frac{x^2e^{1/x}}{x^2}dx$
Let, (going to apply the substitution rule)
$u=-1/x$ then, $u'=1/x^2$
Thus,
$\int \frac{e^u}{u^2}du$
If you use integration by parts with,
$v=e^u$ and $w'=1/u^2$
Then,
$-\frac{e^u}{u}+\int \frac{e^u}{u}du$
We have converted this integral into an integral which has no elementary anti-derivative. This confirms topsquarks guess.

Yeah, what he said.

-Dan

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