# Integral: e^(-1/x)

1. Jan 12, 2010

### Alexx1

Can someone help me with this integral?

e^(-1/x)

2. Jan 12, 2010

### CFDFEAGURU

Just follow the rules for exponential integration.

$$\int$$ $$e^{u}$$ du = $$e^{u}$$ + C

Thanks
Matt

Last edited: Jan 12, 2010
3. Jan 12, 2010

### Staff: Mentor

If this is the integral:
$$\int e^{-1/x}dx$$

an ordinary substitution is not much help. Alexx1, can you show us the complete integral you're trying to do?

4. Jan 12, 2010

### HallsofIvy

Staff Emeritus
If the problem were $\int e^u du$, but it isn't and there is no good way to change it to that form.

It looks to me like $\int e^{1/x} dx$ does not have an anti-derivative in terms of elementary functions.

5. Jan 13, 2010

### g_edgar

in terms of an exponential integral function...

$\int \!{{\rm e}^{-{x}^{-1}}}{dx}=x{{\rm e}^{-{x}^{-1}}}-{\rm Ei}_1 \left({x}^{-1} \right)$

6. Jan 13, 2010

### CFDFEAGURU

HallsofIvy,

Yes, now I see that a simple substitution is not the way to proceed. Thanks for correcting me.

Thanks
Matt