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Homework Help: I need a hand with this integral? probably not even that difficult?

  1. Oct 18, 2009 #1
    I need to evaluate [tex]\int e\stackrel{-(r/R)}{}sin(qr/h)dr[/tex] between r=0 and r=[tex]\infty[/tex] but I keep getting stuck.

    I hae tried integrating by parts but the integral just keeps getting more complicated, and I have tried substituting in sin(x) = (1/2i)(e^ix - e^-ix) but that gave me the result that the whole thing is equal to infinity, which I know to be wrong.

    can anybody suggest some sort of method or substitution that I could use?
    this is driving me up the wall now.
  2. jcsd
  3. Oct 18, 2009 #2
    And yet it should have worked... Did you find an antiderivative for [tex]e^{-\frac r R}\sin\left(\frac{qr}{h}\right)[/tex] ?
  4. Oct 18, 2009 #3


    Staff: Mentor

    Your first approach would have worked if you chose the right things for u and dv, and if you had done integration by parts a second time.

    For the first integration by parts, choose u = e-Ar and dv = cos(Br)dr. For the sake of simplicity, A = 1/R and B = q/h.

    That should give you an integral with e-Ar and sin(Br). Perform integration by parts a second time, with u = e-Ar and dv = sin(Br)dr. This second integration by parts will give you an integral with e-Ar and cos(Br), which is the same as what you started with. At this point you will have your original integral on the left side, and some expression plus a multiple of your original integral on the right side.

    Bring both integral terms to one side and solve algebraically for your integral.
  5. Oct 18, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hi jeebs! :smile:

    (have an integral: ∫ and an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)

    Hint: what is e-r/Rsin(qr/h) the imaginary part of? :wink:
  6. Oct 18, 2009 #5


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    Science Advisor

    Let [itex]u=e^{r/R}[/itex] and [itex]dv= sin(qr/h)dr[/itex]. Then [itex]du= e^{r/R}/R dr[/itex] and [itex]v= -(h/q) cos(qr/h)[/itex].

    [itex]\int e^{r/R}sin(qr/h)dr= -(h/q)e^{r/R}cos(qr/h)+ (h/(qR))\int e^{r/R}cos(qr/h)dr[/itex]. Now do it again, again letting [itex]u= e^{r/R}[/itex] and [itex]dv= sin(qr/h)r[/itex]. You will get a term involving [itex]\int e^{r/R}sin(qr/h)dr[/itex] again. Do not integrate by parts a third time. Instead, solve the equation, algebraically, for [itex]\int e^{r/R}sin(qr/h)dr[/itex].
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