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Does this equality hold, for $\text{k}\in\mathbb{R}$:

$$\mathcal{I}_{\space\text{k}}:=\int_0^\infty\int_0^\infty\sin\left(\text{s}^\text{k}\right)\cdot\cos\left(x^\text{k}\right)\cdot\exp\left(-\text{s}\cdot x\right)\space\text{d}\text{s}\space\text{d}x=$$ $$\int_0^\infty\int_0^\infty\sin\left(\text{s}^\text{k}\right)\cdot\cos\left(x^\text{k}\right)\cdot\exp\left(-\text{s}\cdot x\right)\space\text{d}x\space\text{d}\text{s}\tag1$$

Mathematica 10.0 say that it is correct (for some values of $\text{k}$ I tried). I want to prove that it holds in general, and I've no idea how to start.

I know that Fubini's theorem is a good start, but it leads me to nothing.

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  • $\begingroup$ See Fubini's theorem : en.wikipedia.org/wiki/Fubini%27s_theorem $\endgroup$ – stity Jan 19 '18 at 9:19
  • $\begingroup$ This is exact and true... $\endgroup$ – Mostafa Ayaz Jan 19 '18 at 10:39
  • $\begingroup$ @MostafaAyaz Why? Just saying: "This is exact and true." is not a 'proof'. $\endgroup$ – Jan Jan 19 '18 at 10:41
  • $\begingroup$ This is the same direct application of Fubini's theorem to this problem where the argument is integrable respect to both $x$ and $s$... $\endgroup$ – Mostafa Ayaz Jan 19 '18 at 11:53
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A bit too long for a comment. Here is an idea to show that the integrals are equal, but perhaps you need some convergence argument as well:

The second integral can be written as (just renaming $s$ to $x$ and $x$ to $s$) $$ \int_0^{+\infty}\int_0^{+\infty}\sin(x^k)\cos(s^k)\exp(-sx)\,ds\,dx. $$ Thus, the difference of the integrals equals $$ \int_0^{+\infty}\int_0^{+\infty}\bigl(\sin(s^k)\cos(x^k)-\sin(x^k)\cos(s^k)\bigr)\exp(-sx)\,ds\,dx $$ or, using a trig identity, $$ \int_0^{+\infty}\int_0^{+\infty}\sin(s^k-x^k)\exp(-sx)\,ds\,dx $$ As it happens the sine term is odd in the line $s=x$, so the integral, if there is no issue with convergence, must be $0$.

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