How to find the laplace transform of $\cos(\sqrt t)$?

Question

I tried solving for the transform using the same method the book uses to find laplace transform for $\sin(\sqrt t)$ which is, by writing the Maclaurin's expansion for $\sin(\sqrt t)$ and then using standard Laplace transform for $t^\alpha$ where $\alpha>0$. But the same method doesn't seem to be useful for $\cos(\sqrt t)$. Can someone help ?

Answer 1

Recall that $\cos(\sqrt{t})$ has the hypergeometric series representation $\cos(\sqrt{t})={~}_0F_1(1/2;-t/4)$, where $$ {~}_0F_1(1/2;-t/4)=\sum_{k=0}^\infty \frac{\Gamma(1/2)(-1)^k}{\Gamma(1/2+k)}\frac{t^k}{k!4^k} $$

By employing the Laplace identity $\mathcal{L}\{t^k\}={k!}{s^{-k-1}}$ and the identity $\Gamma(1/2)=\sqrt{\pi}$, there holds $$ \mathcal{L}\{\cos(\sqrt{t})\}(s)=\sum_{k=0}^\infty \frac{\sqrt{\pi}(-1)^k}{\Gamma(1/2+k)4^k}s^{-k-1}=\frac{\sqrt{\pi}}{s}E_{1,1/2}(-1/4s), $$ where $\displaystyle E_{\alpha,\beta}(z)=\sum_{k=0}^\infty \frac{z^k}{\Gamma(\alpha k+\beta)}$ denotes the Mittag-Leffler function.

Alternatively, one can also show that $\mathcal{L}\{\cos(\sqrt{t})\}(s)$ coincides with the hypergeometric series expansion of Kummer type $$ \frac{1}{s}{~}_1F_1(-1,1/2;-1/4s)=\frac{1}{s}\sum_{k=0}^\infty \frac{(-1)_k}{(1/2)_k}\frac{s^{-k}}{k!4^k} $$ showing that $\mathcal{L}\{\cos(\sqrt{t})\}(s)$ is a Laguerre polynomial in disguise -- see, for instance, https://en.wikipedia.org/wiki/Laguerre_polynomials