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I'm struggling to find a closed form for the following distribution (which is after all a Fourier Transform) written in integral form:

$$I=\int_0^\infty\!\!\text{d}k\ \frac{ k }{\sqrt{k^2+m^2}}\sin(k x) \sin \left(t\sqrt{k^2+m^2}\right)$$

where $m,t\in\mathbb{R}$ and $x>0$. I suspect that we will have Dirac deltas and Bessel functions.

Or, equivalently, (and also more practical) if you want to write it in a regularized way as an integral

$$I=\int_0^\infty\!\!\text{d}k\ \frac{ k e^{-\epsilon k}}{\sqrt{k^2+m^2}}\sin(k x) \sin \left(t\sqrt{k^2+m^2}\right)$$

where now it is a convergent regular integral if $\epsilon>0$.

Can anyone help me out? Thanks so much in advance! :)

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A hint for the second integral (since the first doesn't converge).

First let's get rid of $m$ by transforming the variable and the parameters:

$$m\epsilon \to \epsilon,~~~~mx \to x,~~~mt \to t,~~~\frac{k}{m} \to k$$

$$I=m\int_0^{\infty}\frac{k e^{-\epsilon k}}{\sqrt{1+k^2}} \sin (x k) \sin (t \sqrt{1+k^2})dk $$

Now we can use the differentiation under the integral sign to write:

$$I=-m\frac{\partial I_1}{\partial \epsilon}$$

$$I_1=\int_0^{\infty}\frac{e^{-\epsilon k}}{\sqrt{1+k^2}} \sin (x k) \sin (t \sqrt{1+k^2})dk $$

An obvious substitution is:

$$k=\sinh p$$

$$I_1=\int_0^{\infty}e^{-\epsilon \sinh p} \sin (x \sinh p) \sin (t \cosh p)dp $$

We can transform the sine product to the sum of cosines:

$$\sin (x \sinh p) \sin (t \cosh p)=\frac{1}{2} \left(\cos (x \sinh p-t \cosh p)-\cos (x \sinh p+t \cosh p) \right)$$

Then the integral becomes:

$$I_1=\frac{1}{2}\int_0^{\infty}e^{-\epsilon \sinh p} \cos (x \sinh p-t \cosh p) dp- \\ -\frac{1}{2}\int_0^{\infty}e^{-\epsilon \sinh p} \cos (x \sinh p+t \cosh p) dp$$

We can probably use the exponential form:

$$\int_0^{\infty}e^{-\epsilon \sinh p} \cos (x \sinh p-t \cosh p) dp=\Re \left( \int_0^{\infty}e^{-(\epsilon-i x) \sinh p-i t \cosh p}dp \right)$$

Now, the relation to Bessel functions is clear, I will get back and edit the answer if I get closer to solving this.

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