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I have a problem proving $$\int_{0}^\infty dx {\left(\int_{0}^\infty e^{-x^2t}\sin t\, dt\right)}=\int_{0}^\infty dt\left( \int_{0}^\infty e^{-x^2t}\sin t\, dx\right)$$.

I have been struggling for it for a long time, I don't know how to prove it. Or, maybe it is wrong? Can somebody tell me how to prove it? Why they can change the order of integration?

user143444
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python3
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    Some use $\int dt f(t)$ to denote $\int f(t)dt$. Is that the case here? –  May 08 '14 at 23:22
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    Which integrals have you figured out so far? – David H May 08 '14 at 23:23
  • This is an application of Fubini's theorem. You get one side from the other by swapping the order of integration. Since the bounds of the inner integral is fixed in both cases, swapping the order is trivial - you just replace $dx,dt$ by $dt,dx$. Then you'll have to argue that you are in fact allowed to switch the order, i.e. check the prerequisits of Fubini's theorem. – fgp May 08 '14 at 23:26
  • If those brackets are supposed to take care of Sanath Davalapurkar's complaint that your notation is ambiguous - they don't! If anything, they make it worse... – fgp May 08 '14 at 23:28
  • sorry, I haven't learned Fubini's theorem, what is that? – python3 May 08 '14 at 23:33

2 Answers2

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Since you have the integrals specifically, why not just evaluate them? On the LHS

$$\int_0^{\infty} dt \, e^{-x^2 t} \sin{t} = \operatorname{Im} \left [\int_0^{\infty} dt \, e^{-(x^2-i) t} \right ] = \operatorname{Im} \left [\frac1{x^2-i} \right ] = \frac1{x^4+1}$$

$$\int_0^{\infty} \frac{dx}{x^4+1} = \frac{i 2 \pi}{(1-i) (4 e^{i 3 \pi/4})} = \frac{\pi}{2 \sqrt{2}}$$

On the RHS:

$$\int_0^{\infty} dx \, e^{-x^2 t} = \frac12 \sqrt{\frac{\pi}{t}}$$

$$\frac{\sqrt{\pi}}{2} \int_0^{\infty} dt \, \frac{\sin{t}}{\sqrt{t}} = \frac{\sqrt{\pi}}{2}\int_{-\infty}^{\infty} du \, \sin{u^2} = \frac{\pi}{2 \sqrt{2}}$$

Note:

$$\int_{-\infty}^{\infty} du \, \sin{u^2} = 2 \operatorname{Im} \left [\int_{0}^{\infty} du \, \, e^{i u^2} \right ] = 2 \operatorname{Im} \left [e^{i \pi/4}\int_{0}^{\infty} dv \, \, e^{-v^2} \right ] = \frac{\sqrt{\pi}}{\sqrt{2}}$$

So, there you go, you can reverse the order of integration.

Ron Gordon
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  • How about in general? Could you provide a short intro. to Fubini's theorem, for the OP's benefit? (I've done that somewhere, don't remember where.) –  May 08 '14 at 23:37
  • @SanathDevalapurkar: I doubt I could enlighten the OP that way. I am merely answering the question. – Ron Gordon May 08 '14 at 23:38
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    Which is why I gave you a +1, of course. –  May 08 '14 at 23:39
  • can you explain more about figuring out $\int_{-\infty}^{\infty} sinu^{2} du$ – python3 May 08 '14 at 23:48
  • @tiandiao123: I provided a brief explanation. More advanced explanations using Cauchy's theorem are available within the site. – Ron Gordon May 09 '14 at 00:40
  • @RonGordon This is not a proof of the Fresnel integral. You haven't used complex variables. I am not going to down vote because you solve a lot of problems for me. However, this is not complete my friend. – Jeff Faraci May 09 '14 at 00:56
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    @Integrals: guessing the scope of the question, I decided that the hand-waving was the way to go here. Anyhoo, proof embedded within this answer: http://math.stackexchange.com/questions/310942/a-couple-of-fresnel-like-integrals/311182#311182 – Ron Gordon May 09 '14 at 01:08
  • @RonGordon I have seen much better proofs you have posted than that as well. Thanks though. – Jeff Faraci May 09 '14 at 01:12
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Here is a more elementary way of doing this.

\begin{align} \int_{0}^\infty \int_{0}^\infty e^{-x^2t}\sin t \;dt dx&= \int_{0}^\infty \mathcal{L}\{ \sin(t) \} \;dx\\ &=\int_{0}^\infty \frac{1}{(x^2)^2 + 1} dx \quad {\text{You are going to have to use partial fractions and a few substitutions. }}\\ &= \frac{\pi}{2\sqrt{2}} \end{align}

IAmNoOne
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