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I have been trying to use the technique that involves using differentiation under the integral sign but I have had no luck so far. Can you give me some ideas? In this problem the parameter is $t$.

The Pointer
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DouL
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  • Are you allowed to use imaginary numbers? I'd guess the fact that $\sin(tx)$ is the imaginary part of $e^{itx}$ is a useful fact, since we can then look at the integral as the imaginary part of an integral over the complex numbers which is evaluated in an easier way. – Sarvesh Ravichandran Iyer May 13 '21 at 04:14
  • Use the fact that $$x\sin(tx) = -\frac{\partial}{\partial t}\operatorname{Re}\left{e^{itx}\right}$$ for the most efficient solution. Otherwise integration by parts works just fine. – Ninad Munshi May 13 '21 at 04:19
  • Just an idea, why don't you use the Taylor series expansion for $\sin x$, and then you'll get a bunch of gamma functions... – Smriti Sivakumar May 13 '21 at 04:44
  • @Ninad Munshi I don't think integration by parts works here – Ritam_Dasgupta May 13 '21 at 07:27
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    @Ritam It does, integration by parts is always equivalent to differentiating under the integral sign. How can you say that when even the well written, accepted answer uses integration by parts?? – Ninad Munshi May 13 '21 at 07:33
  • @mattos did not just use IBP, other techniques were involved too. If you simply do by parts the second function you get is not integrable in closed form. – Ritam_Dasgupta May 13 '21 at 07:39

1 Answers1

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If you want to use differentiation under the integral sign, maybe first consider the integral

$$I(c,t) = - \int_{0}^{\infty} e^{-cx^{2}} \cos(xt) dx$$

where we are assuming $c > 0$. As the integrand meets the conditions for differentiating under the integral, then doing so and using integration by parts we find

\begin{align} I_{t}(c,t) &= \int_{0}^{\infty} x e^{-cx^{2}} \sin(xt) dx \\ &= - \frac{1}{2c} \int_{0}^{\infty} (-2 c x e^{-cx^{2}}) \sin(xt) dx \\ &= - \frac{1}{2c} \int_{0}^{\infty} (e^{-cx^{2}})' \sin(xt) dx \\ &= - \frac{1}{2c} \left[ \sin(xt) e^{-cx^{2}} \bigg \lvert_{0}^{\infty} - \int_{0}^{\infty} t e^{-cx^{2}} \cos(xt) dx \right] \\ &= - \frac{t}{2c} I(c,t) \\ \end{align}

which you can then solve to get

$$I = A e^{-t^{2}/4c}$$

Considering $I(1,0) = A = - \sqrt{\pi}/2$ using this, we have

$$I = - \frac{\sqrt{\pi}}{2} e^{-t^{2}/4c}$$

and so

$$I_{t} = \int_{0}^{\infty} x e^{-cx^{2}} \sin(xt) dx = \frac{\sqrt{\pi}}{2} \cdot \frac{t}{2c} e^{-t^{2}/4c}$$

Matthew Cassell
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