What am I doing wrong in this simple exponential integral?

[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?

 

[Dickfore]

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

 

[exitwound]

In English would be preferable. :)

 

[Dickfore]

In English would be preferable. :)

Please cover the section on Parametric Integrals. kthnxbai.

 

[D H]

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

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.

 

[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.

 

[Dickfore]

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.

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!

 

[D H]

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.

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.

 

[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?

 

[Dickfore]

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?

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.

 

[exitwound]

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