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I encountered this integral in my Quantum Mechanics class archives (which was apparently asked during an oral exam in the past years) : $t \in \mathbb{R}$ is a parameter

$$I(t) = \int^{\frac{1}{2}}_{-\frac{1}{2}} \sin(t(x^2-x^4))dx$$

To be clear there was no context given and I can't figure out how it would relate to any other problem so I think a mathematical approach would be best.

(I tried multiple things including derivating the whole thing multiple times but had no significant advancement, I also don't expect people to solve it for me but any idea would be welcome.)

gazo
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  • Quantic physics? Pretty sure you can’t simplify that integral. What do you need to do with it? – Eric Dec 29 '23 at 15:26
  • Quantum Mechanics* , my problem is that there is absolutely 0 context and now I'm so frustrated that I can't figure out how to solve it – gazo Dec 29 '23 at 15:30
  • I'm not sure there's a way to solve that problem, as you can see here. – Accelerator Dec 29 '23 at 16:25

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As a double infinte series. $$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sin \left(t \left(x^2-x^4\right)\right) \, dx=\\\Im\left(\int_{-\frac{1}{2}}^{\frac{1}{2}} \exp \left(i t \left(x^2-x^4\right)\right) \, dx\right)=\\\Im\left(\int_{-\frac{1}{2}}^{\frac{1}{2}} \left(\sum _{n=0}^{\infty } \frac{\left(i t x^2\right)^n}{n!}\right) \sum _{m=0}^{\infty } \frac{\left(-i t x^4\right)^m}{m!} \, dx\right)=\\\Im\left(\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\left(i t x^2\right)^n \left(-i t x^4\right)^m}{n! m!} \, dx\right)=\\\Im\left(\sum _{n=0}^{\infty } \sum _{m=0}^{\infty } \frac{2^{-2 (2 m+n)} (-i t)^m (i t)^n}{(1+4 m+2 n) m! n!}\right)$$

Maybe can be or not expressed by Kampé de Fériet function.

Mariusz Iwaniuk
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