What am I doing wrong in this simple exponential integral?

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exitwound
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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:

[tex]\int_0^\infty{e^{-xy}dy}[/tex]

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 [itex]e^{-xy}[/itex] is [itex]-xe^{-xy}[/itex] correct?
 
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Treat [tex]x[/tex] as a constant when doing this table integral with respect to the dummy variable [tex]y[/tex]. It only converges if [tex]x > 0[/tex].
 
In English would be preferable. :)
 
exitwound said:
In English would be preferable. :)

Please cover the section on Parametric Integrals. kthnxbai.
 
exitwound said:
The antiderivative of [itex]e^{-xy}[/itex] is [itex]-xe^{-xy}[/itex] correct?
That is incorrect. Differentiate your result with respect to y. Do you get [itex]e^{-xy}[/itex]?

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

[tex]\int_0^{\infty} e^{-ax}\,dx[/tex]

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.
 
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:

[tex]\int_0^{\infty} e^{-ax}\,dx[/tex]

[tex](\frac{e^{-ax}}{-a})_0^\infty[/tex]

[tex](0-(1/-a))=1/a[/tex]

yes?
 
exitwound said:
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:

[tex]\int_0^{\infty} e^{-ax}\,dx[/tex]

[tex](\frac{e^{-ax}}{-a})_0^\infty[/tex]

[tex](0-(1/-a))=1/a[/tex]

yes?

Now, substitute [tex]a \rightarrow x[/tex] 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.
 
So:
[tex]\int_0^\infty{e^{-xy}}dy = \frac{e^{-xy}}{-x}_0^\infty = 0+1/x[/tex]