Find Taylor series expansion and convergence radius for $\int_0^x\cos(\sqrt{t}\ )dt$ i must find the the Taylor series expansion (i've been asked not necessarily calculating it directly) and the convergence radios for this function : 
$$f(x) = \int_0^x \cos(\sqrt{t}\ ) \, dt$$
I am new to this field, and im not really sure what do i need to do, 
so maybe this is an elementary question, but i'd appreciate it if you add explanations so i can understand.
Thanks alot.
 A: Hint: to find the Taylor series expansion for this function, find the Taylor series expansion for $\cos(\sqrt{t})$, and then integrate it from $0$ to $x$. 
A: Let $u=u(t)=\sqrt{t}\,$. Then
\begin{align*}
\bigl(\cos\sqrt{t}\,\bigr)^{(n)}
&=\sum_{k=0}^n\cos^{(k)}u B_{n,k}\biggl(\biggl\langle\frac12\biggr\rangle_1 t^{-1/2}, \biggl\langle\frac12\biggr\rangle_2 t^{-3/2}, \dotsc, \biggl\langle\frac12\biggr\rangle_{n-k+1}t^{1/2-(n-k+1)}\biggr)\\
&=\sum_{k=0}^n\cos\biggl(u+\frac{k\pi}{2}\biggr) \bigl(\sqrt{t}\,\bigr)^{k}\frac{1}{t^n}B_{n,k}\biggl(\biggl\langle\frac12\biggr\rangle_1, \biggl\langle\frac12\biggr\rangle_2, \dotsc, \biggl\langle\frac12\biggr\rangle_{n-k+1}\biggr)\\
&=\sum_{k=0}^n\cos\biggl(\sqrt{t}\,+\frac{k\pi}{2}\biggr) \bigl(\sqrt{t}\,\bigr)^{k}\frac{1}{t^n}(-1)^k\frac{n!}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{\ell/2}{n}\\
&=\frac{n!}{t^n}\sum_{k=0}^n\frac{(-1)^k}{k!}\cos\biggl(\sqrt{t}\,+\frac{k\pi}{2}\biggr) t^{k/2}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{\ell/2}{n}.
\end{align*}
From this general formula for the $n$th derivative of $\cos\sqrt{t}\,$, one can derive Maclaurin's series expansions of the functions $\cos\sqrt{t}\,$ and $f(x)=\int_0^x\cos\sqrt{t}\,\text{d}t$.
Reference

*

*Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.

Alternatively and simply, it is common knowledge that
$$
\cos t=\sum_{k=0}^\infty(-1)^k\frac{t^{2k}}{(2k)!},\quad |t|<\infty.
$$
Therefore, we have
$$
\cos \sqrt{t}\,=\sum_{k=0}^\infty(-1)^k\frac{t^{k}}{(2k)!},\quad 0\le t<\infty.
$$
Consequently, it follows that
$$
\int_0^x\cos \sqrt{t}\,\text{d}t=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\frac{t^{k+1}}{k+1},\quad 0\le t<\infty.
$$
