How to apply Laplace transform to integral? Question: For a real number x, assuming $f(t)=\int_0^\infty \frac{\sin{tx}}{\sqrt{x}}dx$, apply Laplacetransformation to $f(t)$ and find $f(t)$.
Through Laplace transform, I can find $\int_0^\infty\frac{1}{\sqrt{x}}\int_0^\infty\sin{(tx)}e^{-st}dtdx=\int_0^\infty\frac{1}{\sqrt{x}}\frac{x}{s^2+x^2}dx$, then I couldn't find a way to continue this. Did I made a mistakes here?
 A: Well, we are trying to find the following integral:
$$\mathcal{I}_\text{n}\left(\text{k}\right):=\int_0^\infty\frac{\sin\left(\text{n}x\right)}{x^\text{k}}\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
$$\mathcal{I}_\text{n}\left(\text{k}\right)=\int_0^\infty\mathscr{L}_x\left[\sin\left(\text{n}x\right)\right]_{\left(\sigma\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{k}}\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$
Which gives:
$$\mathcal{I}_\text{n}\left(\text{k}\right)=\int_0^\infty\frac{\text{n}}{\text{n}^2+\sigma^2}\cdot\frac{\sigma^{\text{k}-1}}{\Gamma\left(\text{k}\right)}\space\text{d}\sigma=\frac{\text{n}}{\Gamma\left(\text{k}\right)}\int_0^\infty\frac{\sigma^{\text{k}-1}}{\text{n}^2+\sigma^2}\space\text{d}\sigma\tag3$$
Using this answer, we can see that:
$$\mathcal{I}_\text{n}\left(\text{k}\right)=\frac{\pi\text{n}^{\text{k}-1}}{2\Gamma\left(\text{k}\right)}\cdot\csc\left(\frac{\text{k}\pi}{2}\right)\tag4$$

So using $\text{n}=t$ and $\text{k}=\frac{1}{2}$, we get:
$$\mathcal{I}_t\left(\frac{1}{2}\right)=\frac{\pi t^{\frac{1}{2}-1}}{2\Gamma\left(\frac{1}{2}\right)}\cdot\csc\left(\frac{1}{2}\cdot\frac{\pi}{2}\right)=\frac{1}{\sqrt{t}}\cdot\frac{\sqrt{\pi}}{\sqrt{2}}=\sqrt{\frac{\pi}{2t}}\tag5$$

