<s>[INST] Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$.
Remember that
$J_k(z)=c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz$ where $c_k=\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}$
$J_{k-1}(z)=c_{k-1} z^{k-1} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{3}{2}}dz$ where $c_{k-1}=\frac{1}{2^{k-1} \Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}$
$$\frac{d}{dz}(z^kJ_k(z)) =\frac{d}{dz}(z^k c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k \frac{d}{dz}(z^{2k} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k [2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)+ z^{2k} (e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1] $$
But $e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1=0$. So 
$$\frac{d}{dz}(z^kJ_k(z)) =c_k 2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz) =\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =\frac{1}{2^k-1 (k-\frac{1}{2})\Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}kz^{k-1}z^k\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =z^k c_{k-1}kz^{k-1}\int_{-1}^1 \frac{e^{izs}(1-z^2)^{k-\frac{1}{2}}}{k-\frac{1}{2}}dz$$
How do I contunie from here?
Write an explicit expression for $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$
We know that $J_\frac{1}{2}(x)=\sqrt{\frac{2}{\pi}}\frac{\sin(x)}{\sqrt{x}}$ when $c_\frac{1}{2}=\frac{1}{\sqrt{2}\Gamma(1)\Gamma(\frac{1}{2})}$. By the formula in the previous part, we now that 
$$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x) \iff x^kJ_k(x)=\int_\textbf{R} x^kJ_{k-1}(x)dx \iff J_k(x)=\int_\textbf{R} J_{k-1}(x)dz$$
$$J_\frac{3}{2}(z) =\int_\textbf{R}J_\frac{1}{2}(x) \,dx= \sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx =\sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx=(1+i)\frac{\pi}{2}$$
$$J_\frac{5}{2}(x) =\int_\textbf{R}J_\frac{3}{2}(x) \,dx =\int_\textbf{R}(1+i){\frac{\pi}{2}} \,dx $$
Does $J_\frac{5}{2}(x)=\infty$?
Write an explicit expression for $J_1+2J_3$ in terms of $J_0$
By the formula in the previous part,
$$J_1(x) =\int_\textbf{R} J_0(x)\,dx$$
$$J_3(x) =\int_\textbf{R}J_2(x)=\int_{\textbf{R}^2} J_1(x)\,dx^2=\int_{\textbf{R}^3} J_0(x)\,dx^3 $$
So $J_1+2J_3=\int_\textbf{R} J_0(x)\,dx+2\int_{\textbf{R}^3} J_0(x)\,dx^3$
Please check
 [/INST] You have a very bad habit of repeating variables when you should not. This is a grave mistake and causes errors like you have made. I'm going to get a series expansion for the Bessel function because the integrand is fairly difficult to work with, in my opinion. Series are quite nice though-especially with respect to differentiation! By your definition,
$$J_{\nu}(z) = c_{\nu}z^{\nu}\int_{-1}^1 e^{izt}(1-t^2)^{\nu-\frac{1}{2}}\,dt,$$
where $c_{\nu} = \dfrac{1}{\pi^{\frac{1}{2}}2^{\nu}\, \Gamma\left(\nu+\frac{1}{2}\right)}.$ Concentrating on the case where $z$ is real, this becomes
$$J_{\nu}(x) = c_{\nu}x^{\nu}\int_{-1}^1 e^{ixt}(1-t^2)^{\nu-\frac{1}{2}}\,dt.\tag{1}$$
The nonlinearity of this integrand makes it difficult to work with but we can get around this by noting two things: firstly, we needn't consider the imaginary part due to symmetry and by expanding $\cos(xt)$ in a power series. We can interchange integral and summation since the convergence of the power series is uniform.
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_{-1}^1 t^{2n}(1-t^2)^{\nu-\frac{1}{2}}\,dt.$$
Making use of evenness,
$$J_{\nu}(x) = 2c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_0^1 t^{2n}(1-t^2)^{\nu-\frac{1}{2}}\,dt.$$
Let's make the change of variable $y = t^2$ then $dy = 2t\,dt$ which gives
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\int_0^1 y^{n-\frac{1}{2}}(1-y)^{\nu-\frac{1}{2}}\,dy.$$
Noting the relationship this integral has to the Gamma function, we get that
$$J_{\nu}(x) = c_{\nu}x^{\nu}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\frac{\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\nu+\frac{1}{2}\right)}{\Gamma(n+\nu+1)}.$$
We almost have the right expression at our disposal now so that we can easily analyze these problems. We need only to simplify $\Gamma\left(n+\frac{1}{2}\right)$. This can be rewritten as
$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)!\pi^{\frac{1}{2}}}{4^nn!}.$$
(See here for more information.) Thus we get a nice expression for $J_{\nu}$:
$$J_{\nu}(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+\nu}\Gamma(n+\nu+1)n!}x^{2n+\nu}.$$
This series converges for all $x$ and so we can differentiate it term-by-term. If $\nu=m$, multiplying by $x^m$ gives
$$(x^mJ_m(x))' = \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+m}\Gamma(n+m+1)n!}(x^{2n+2m})'.$$
$(x^{2n+2m})' = (2n+2m)x^{2n+2m-1} = 2(n+m)x^m(x^{2n+(m-1)})$. This gives
$$(x^mJ_m(x))' = x^m\sum_{n=0}^{\infty}\frac{(-1)^n2(n+m)}{2^{2n+m}\Gamma(n+m+1)n!}x^{2n+(m-1)}.$$
Making use of the fact that $\Gamma(x+1) = x\Gamma(x)$, we get that
$$(x^mJ_m(x))' = x^m\sum_{n=0}^{\infty}\frac{(-1)^n(n+m)}{2^{2n+m-1}(n+m)\Gamma(n+m)n!}x^{2n+(m-1)} = x^m J_{m-1}(x).$$
You don't have to restrict to the real case if you are not so inclined. The work above is quite general but easiest to frame in the real-line case.
With this in mind, we have $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}\sin x.$
Therefore by the above, 
$$\left(x^{\frac{3}{2}}J_{\frac{3}{2}}(x)\right)' = x^{\frac{3}{2}}J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi}}x\sin x.$$
Integrating both sides gives 
$$x^{\frac{3}{2}} J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi}} \int x\sin x\,dx = \sqrt{\frac{2}{\pi}}(\sin x-x\cos x)+C.$$
Setting $x=0$ gives $C=0$ and so
$$J_{\frac{3}{2}}(x) = \sqrt{\frac{2}{\pi x^3}}(\sin x-x\cos x).$$
We employ a very similar procedure for computing $J_{\frac{5}{2}}$:
$$\left(x^{\frac{5}{2}} J_{\frac{5}{2}}(x)\right)' = x^{\frac{5}{2}} J_{\frac{3}{2}}(x).$$
Integrating both sides gives
$$x^{\frac{5}{2}} J_{\frac{5}{2}}(x) = \sqrt{\frac{2}{\pi}}\int x(\sin x-x\cos x) = \sqrt{\frac{2}{\pi}} (-x^2\sin x -3x\cos x+3\sin x)+C.$$
Again, setting $x=0$ gives $C=0$ and so
$$J_{\frac{5}{2}}(x) = \sqrt{\frac{2}{\pi x^5}}(-x^2\sin x - 3x\cos x+3\sin x).$$
I trust that you can take it from here as the remaining case (part (b)) is quite similar in nature.
</s>