# nasty integral

• Sep 7th 2006, 03:04 PM
v71
nasty integral
which is the integral for exp( -a/x ) dx ?

i have tried every method

thanks
• Sep 7th 2006, 04:11 PM
topsquark
Quote:

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
• Sep 7th 2006, 05:34 PM
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.
• Sep 7th 2006, 05:44 PM
topsquark
Quote:

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