I'm trying to evaluate the following integral;

$$\int e^{(x^2 - z^2)} (2x \cos(2xz) - 2z \sin(2xz)) dz$$

I've tried splitting it up, and using integration by parts, but it just isn't coming out in a simple way. I've been stuck on this for hours. I'm sure there's some rule or trick I can use, but I'm really not sure.

Any assistance would be fantastic. :)

  • $\begingroup$ Nope. I've just gotta integrate that with respect to z. $\endgroup$ – Jack Oct 16 '13 at 20:20
  • $\begingroup$ The second is related to the first trough a derivative respect $x$. $\endgroup$ – Felix Marin Oct 19 '13 at 12:52

Hint: Observe that the first $e^{x^2}$ factors out of the integral. After removing it, notice that what you have looks sort of like the output of a product rule. Can you work backwards to find the product?

  • $\begingroup$ Hm, mine would be faster but I just happened to see it; Andrew's is probably better. $\endgroup$ – Eric Stucky Oct 16 '13 at 20:30
  • $\begingroup$ Ah, I'm rather sleep-deprived. I'm not really seeing it, sorry. $\endgroup$ – Jack Oct 16 '13 at 20:37
  • $\begingroup$ Ahhh, wait a minute. If we factor out both $e^{x^2}$ and $2$, we can set $f(x) = cos(2xz)$, and $g(x) = z e^{-z^2}$, (such that $g'(x) = (1 - 2z^2) e^{-z^2}$, setting $x = 1 - 2z^2$), right?? $\endgroup$ – Jack Oct 16 '13 at 20:48
  • $\begingroup$ @Jack: Very close, but not quite. I don't know where your $f$ went, and also you don't want any $z^2$ terms anywhere except in the exponential. $\endgroup$ – Eric Stucky Oct 16 '13 at 21:10

Hint: Distribute the product and split into two integrals. Do nothing with the first, and integrate the second by parts, using $u = \sin(2xz)$.

  • $\begingroup$ So, by splitting into two functions, you mean, I should end up with something like, say, $$\int e^{(x^2 - z^2)} 2x \cos(2xz) dz$$ and $$-\int e^{(x^2 - z^2)} 2z \sin(2xz) dz$$ However, how does the substitution for $u = \sin(2xz)$?? Does $u$ just become a constant, or do I treat it as a function of $z$?? $\endgroup$ – Jack Oct 16 '13 at 20:35
  • $\begingroup$ @Jack: Yes, split up as you wrote, and integrate by parts with the suggested $u$. :) $\endgroup$ – Andrew D. Hwang Oct 16 '13 at 20:47
  • $\begingroup$ So, if I let $u = \sin(2xz)$, I can get that $du = -\frac{1}{2x} cos(2xz)$, but I'm not sure how to put that into the second?? $\endgroup$ – Jack Oct 16 '13 at 20:52
  • $\begingroup$ Almost; check your usage of the chain rule. :) $\endgroup$ – Andrew D. Hwang Oct 16 '13 at 20:54
  • $\begingroup$ Ah, obviously. So, I should actually get $du = \frac{-1}{2x} \cos(2xz) dz$. From there, I just sub in for du and u, correct?? $\endgroup$ – Jack Oct 16 '13 at 21:12

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