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

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 ?

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