# Simple exponential integral

1. Apr 18, 2010

### exitwound

This is part of double integral. I just can't seem to figure out what I'm doing wrong.

The inner integral comes out to be:

$$\int_0^\infty{e^{-xy}dy}$$

I emailed my teacher to help me through it, and he says this should integrate down to 1/X but I can't seem to figure out how. I'm no good with exponentials and differentiation/integration.

The antiderivative of $e^{-xy}$ is $-xe^{-xy}$ correct?

2. Apr 18, 2010

### Dickfore

Treat $$x$$ as a constant when doing this table integral with respect to the dummy variable $$y$$. It only converges if $$x > 0$$.

3. Apr 18, 2010

### exitwound

In English would be preferable. :)

4. Apr 18, 2010

### Dickfore

Please cover the section on Parametric Integrals. kthnxbai.

5. Apr 18, 2010

### D H

Staff Emeritus
That is incorrect. Differentiate your result with respect to y. Do you get $e^{-xy}$?

Get in the habit of double-checking your work, and in longer calculations, double-check your intermediate results. It might take a bit of extra work to do so, but it will save a lot of wasted effort.

6. Apr 18, 2010

### exitwound

Obviously, I'm not sure what I'm doing wrong. I know that down to this point, everything's correct, as I've double-checked it vs what the teacher had given to me.

I realize I have to differentiate it with respect to y but this is where I'm having trouble. Would it be $(e^{-xy})/-x$? I'm not sure how to go through this one.

7. Apr 18, 2010

### Dickfore

it is a definite integral! You need to use the newton - leibinitz formula (a.k.a the fundamental theorem of calculus) and substitute the limits!

8. Apr 18, 2010

### D H

Staff Emeritus
You are making this too hard. This is an easy problem.

Suppose a is some constant. What is

$$\int_0^{\infty} e^{-ax}\,dx$$

Conceptually, there is zero difference between the above and the problem at hand. Stop thinking of x as always being a variable. It isn't in this case.

9. Apr 18, 2010

### exitwound

That's what I've been trying to do for a few hours now. I know x isn't a variable. I know it's a constant.

In the case of your question:

$$\int_0^{\infty} e^{-ax}\,dx$$

$$(\frac{e^{-ax}}{-a})_0^\infty$$

$$(0-(1/-a))=1/a$$

yes?

10. Apr 18, 2010

### Dickfore

Now, substitute $$a \rightarrow x$$ and look at the hint your instructor gave you. Also, consider when you can make the value on the upper bound equal zero as you did.

11. Apr 18, 2010

### exitwound

So:
$$\int_0^\infty{e^{-xy}}dy = \frac{e^{-xy}}{-x}_0^\infty = 0+1/x$$