I have just started Integral transformations and right now I know very basic things about Laplace and Fourier transformations. While surfing through YouTube, I came across a video:
Find the Laplace transformation of $\frac{\cos\sqrt t}{\sqrt t},t>0.$
But I'm not aware of the properties that he (the YT guy) used to evaluate it, yet. The problem seemed interesting to me so I decided to give it a try (using the definition of Laplace transformation).
My work so far:
We have:
$$\mathcal L\left\{\frac{\cos\sqrt t}{\sqrt t}\right\}=\int_0^{\infty} e^{-st}\cdot\frac{\cos\sqrt t}{\sqrt t}\,dt\\\\=\int_0^{\infty}e^{-st}\cdot\frac{e^{i\sqrt t}+e^{-i\sqrt t}}{2\sqrt t}\,dt\\\\=\int_0^{\infty}e^{-su^2}(e^{iu}+e^{-iu})\,du,\text{ via the substitution }\sqrt t=u\\\\=\int_0^{\infty}\left(e^{-s(u^2-iu/s)}+e^{-s(u^2+iu/s)}\right)\,du\\\\=e^{\frac{-1}{4s}}\int_0^{\infty}\left[\exp\left(-s\left(u-\frac i{2s}\right)^2\right)+\exp\left(-s\left(u+\frac i{2s}\right)^2\right)\right]\,du,\text{ via completing the squares}\\\\=\frac{e^{\frac{-1}{4s}}}{\sqrt s}\left[\int_{-\frac i{2\sqrt s}}^{\infty}e^{-v^2}\,dv+\int_{\frac i{2\sqrt s}}^{\infty}e^{-w^2}\,dw\right],\text{ via the substitutions }v=\sqrt s\left(u-\frac i{2s}\right)\text{ and }w=\sqrt s\left(u+\frac i{2s}\right).$$ I got stuck here due to the lower limits of the integrals above.
The answer to this question is: $\sqrt{\fracπs}\cdot e^{-\frac1{4s}}$.
As far as I know we cannot determine whether $\frac i{2\sqrt s}$ is positive or not. But if I consider: $\frac i{2\sqrt s}>0.$ Then we can easily see that the required Laplace transform is given by:$$\frac{e^{-\frac1{4s}}}{\sqrt s}\left[\int_0^{\infty}e^{-v^2}\,dv+\int_0^{\infty}e^{-w^2}\, dw\right].$$ Both the above integrals are equal and can be evaluated by substituting either $v^2=x$ or $w^2=x.$ Then we have: \begin{align} \mathcal L\left\{\frac{\cos\sqrt t}{\sqrt t}\right\}&=\frac{e^{-\frac1{4s}}}{\sqrt s}\cdot2\cdot\frac{\Gamma\left(\frac12\right)}2\\\\&=\frac{e^{-\frac1{4s}}}{\sqrt s}\cdot\sqrtπ\\\\&=\sqrt{\fracπs}\cdot e^{-\frac1{4s}}. \end{align} Which matches with the answer.
Please suggest, how to tackle the lower bounds or how to modify my method?
(Sorry for the tittle. Couldn't think of a proper one..)