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Simple exponential integral

  1. Apr 18, 2010 #1
    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?
     
  2. jcsd
  3. Apr 18, 2010 #2
    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].
     
  4. Apr 18, 2010 #3
    In English would be preferable. :)
     
  5. Apr 18, 2010 #4
    Please cover the section on Parametric Integrals. kthnxbai.
     
  6. Apr 18, 2010 #5

    D H

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    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.
     
  7. Apr 18, 2010 #6
    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.
     
  8. Apr 18, 2010 #7
    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!
     
  9. Apr 18, 2010 #8

    D H

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    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.
     
  10. Apr 18, 2010 #9
    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?
     
  11. Apr 18, 2010 #10
    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.
     
  12. Apr 18, 2010 #11
    So:
    [tex]\int_0^\infty{e^{-xy}}dy = \frac{e^{-xy}}{-x}_0^\infty = 0+1/x [/tex]
     
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