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I've been at this for a few hours now, and it's frighteningly similar to the problem stated here:

How to prove $\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$

but with enough change that it's still proving problematic. I also had some parts of the solution fly right over my head! I've been trying to differentiate w.r.t. b and find a clever u-sub to no avail. I'm also curious of using Euler's identity to exchange the cos for some exponentials, but I keep hitting a wall there as well. Help on either front would be great.

Edit: a>0 and b>0 are arbitrary constants just to be complete about this.

Update: Solved thanks to voldemort's solution and the Differential Equation solution from the linked problem, thanks all!

Community
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slamDuncan
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  • Do as you mentioned use the exponential form for the cosine and then you shall complete the square. After that change of variables and integrate and only look at the real part :) – Chinny84 Sep 03 '14 at 23:19
  • The answers given are good. Where is it you are getting stuck? If you show what you are doing, we can determine what is giving you trouble. – robjohn Sep 03 '14 at 23:35

2 Answers2

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In your integral, $a$ is presumably positive, else we would not have convergence.

Make the change of variable $t=x\sqrt{a}$ and you will be at an integral very close to the one linked to.

André Nicolas
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Just substitute $u^2=ax^2$, and use the result of the previous integration.

So, we must have $a >0$ (convergence issues), and thus, $u=\sqrt a x$, and $du=\sqrt a dx$.

After this substution, your integrand becomes $\frac{1}{\sqrt a} (e^{-u^2}\cos(\frac{b}{\sqrt a}u) du)$, and this is just the integral you linked to, with $\frac{b}{\sqrt a}$ instead of $2b$.

Remember that $x,u,b$ etc are just "variables".

Previous integral is also same as $\int_{0}^{\infty}e^{-y^2}\cos(b_1y)dy=\frac{\sqrt{\pi}}{2}e^{-b_1^2/4}$- hopefully it will be easier to make the connection now, by taking $b_1=\frac{b}{\sqrt a}$.

voldemort
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  • Could you elaborate/be more specific? I'm trying your suggestion to no avail, but I think that's because I'm missing the gist of it. – slamDuncan Sep 03 '14 at 23:30
  • @slamDuncan: Let me edit my answer. – voldemort Sep 03 '14 at 23:32
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    I was blundering, alright, this makes more sense now. I've made it to the bit in the other problem where the differential is changed from dx (du here obv.) to de^(-x^2) and I'm baffled, I think this is the last hurdle though. – slamDuncan Sep 03 '14 at 23:44
  • @robjohn Yes, this answer is sufficient, my issues are now as listed in my previous comment. – slamDuncan Sep 03 '14 at 23:51
  • @slamDuncan: edited a bit more for clarity. Please let me know if it helps. – voldemort Sep 03 '14 at 23:52
  • The addition is clear, I've come that far. I need more than just the eventual answer though. Once I can get to d/db f(b) = -2b f(b) (Listed in the linked problem, obviously slightly different as analogous for these constants) I'm fine, but I'm unsure how to reach such an equivalence. – slamDuncan Sep 03 '14 at 23:57
  • @slamDuncan: you do not need to differentiate wrt $b$- treat it as a "constant"- but think of it as a symbol- so that you can "plug in" different values. I am not sure if I am confusing you more or helping.. but think with some specific values of $b$ like say $b=2$ to see my point. – voldemort Sep 03 '14 at 23:59
  • I'm sorry, it only confused. I was under the impression that the solution to this would involve differentiating wrt b, doing algebra to create the original integral somehow, and setting this up as an easily solvable ODE which can then be evaluated at some point to find proper constants. However, I seem to be missing a critical step here. – slamDuncan Sep 04 '14 at 00:05
  • @slamDuncan: We are asking you to change your integral into the other form- so that you can use the other integral's result... so a proof would go like this:
    1. Prove the previous result (which is done in the link provided)
    2. By making suitable change of variables- as in my answer, reduce your integral to the integral in step 1)
    3. Thus you get your answer.
     – voldemort Sep 04 '14 at 00:09
  • Yes! This is good, but I don't understand how the previous result works, specifically the step where e^(-x^2)2xsin(2bx)dx is transformed into sin(2bx)de^(-x^2) and then back. Everything else follows. – slamDuncan Sep 04 '14 at 00:12
  • @slamDuncan: ah! That was integration by parts in the previous answer- is that what your problem was? – voldemort Sep 04 '14 at 00:13
  • That's it! It's embarrassing to admit, but I'm having trouble doing it in this problem, if you'd be willing to help there too. I greatly appreciate this. – slamDuncan Sep 04 '14 at 00:17
  • @slamDuncan: Doing it in this problem is a bit nastier- just a tad bit... Usually if we have a previous result that we have proved, we use it to do our case :-P I shall return and edit the answer a bit later- if you haven't figured it out... I have to leave now. I shall be happy to help later- hope that fine with you :). – voldemort Sep 04 '14 at 00:22
  • It's fine, I'll do what I can, but I look forward to your solution! Thanks again. – slamDuncan Sep 04 '14 at 00:23