Variant 1: Harmonic functions have the mean value property,
$$f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\varphi})\,d\varphi$$
if $f$ is harmonic in $\Omega$ and $\overline{D_r(z)} \subset \Omega$.
$u$ is an entire harmonic function, hence
$$u(0) = \frac{1}{2\pi}\int_0^{2\pi} u(e^{i\varphi})\,d\varphi = \frac{1}{2\pi}\int_0^{2\pi} \cos(\varphi)e^{\cos\varphi}\cos (\sin\varphi) - \sin(\varphi)e^{\cos\varphi}\sin(\sin\varphi)\,d\varphi.$$
Whether we write $z$ and $re^{i\varphi}$ or $(x,y)$ and $(r\cos \varphi, r\sin\varphi)$ is completely immaterial. The complex notation is just more convenient sometimes.
Variant 2: Consider the analytic function $f = u+iv$.
Since $u$ and $v$ are both real, and $d\varphi$ is also real, we have
$$\begin{align}
\int_0^{2\pi} u(\cos\varphi,\sin\varphi)\,d\varphi &= \operatorname{Re}\left(\int_0^{2\pi} u(\cos\varphi,\sin\varphi)\,d\varphi + i\int_0^{2\pi} v(\cos\varphi,\sin\varphi)\,d\varphi\right)\\
&= \operatorname{Re} \int_0^{2\pi} f(e^{i\varphi})\,d\varphi.
\end{align}$$
Now,
The path integral over some closed curve is zero, over an analytic function.
is not correct as stated. On the one hand, the closed curve must not wind around any point in the complement of the function's domain - but since we have an entire function, that is vacuously satisfied here. More pertinent in the case at hand is that the integral theorem concerns only integrals with respect to $dz$ (it's a theorem about holomorphic differential forms), but here the integrand is $f(z)\,d\varphi$, not $f(z)\,dz$. Thus Cauchy's integral theorem does not apply.
However, for integrals over a circle, we have a simple correspondence between $dz$ and $d\varphi$. If we parametrise the circle as $\gamma(\varphi) = z_0 + r e^{i\varphi}$, then we have
$$dz = \gamma'(\varphi)\,d\varphi = ire^{i\varphi}\,d\varphi = i(z-z_0)\,d\varphi,$$
so we get
$$\int_0^{2\pi} f(e^{i\varphi})\,d\varphi = \int_{\lvert z\rvert = 1} f(z)\frac{dz}{iz},$$
and we see that that leads to Cauchy's integral formula,
$$\frac{1}{i} \int_{\lvert z\rvert = 1} \frac{f(z)}{z}\,dz = 2\pi\: f(0).$$
We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$
(See this question for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases}
\frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\
\frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0
\end{cases} $$
Best Answer
By using the exponential generating function for the Legendre polynomials: \begin{equation} e^{xz}J_{0}\left(z\sqrt{1-x^{2}}\right)=\sum_{n=0}^{\infty}\frac{P_{n}\left(x\right)}{n!}z^{n} \end{equation} with $z=-t,x=\cos\theta$ the first integral (taken with $a=1$, as remarked by @Gary in a comment) reads \begin{align} \varphi (R) &= \int_0^\frac{\pi}{2} \int_0^R e^{-t\cos\theta} \sin^3\theta\, J_0(t\sin\theta) \, dt \, d\theta \\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^\frac{\pi}{2} \int_0^Rt^nP_n(\cos\theta)\sin^3\theta \, dt \, d\theta\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!}R^{n+1}\int_0^1(1-x^2)P_n(x)\,dx \end{align} we supposed valid the interchange of integration and summation and change $x=\cos\theta$ to obtain the latter expression.
When $n$ is even ($n=2p$), we have, by symmetry, \begin{align} w_{2p}&=\int_0^1(1-x^2)P_{2p}(x)\,dx\\ &=\frac{1}{2}\int_{-1}^1(1-x^2)P_{2p}(x)\,dx \end{align} The orthogonality properties of the Legendre polynomials indicate that the integrals with $p\ge2$ vanish and direct calculation shows that $w_0=2/3,w_2=-2/15$.
When $n$ is odd ($n=2p+1$), we use the tabulated integral (DLMF) \begin{equation} \int_{0}^{1}P_{2p+1}\left(x\right)x^{z-1}\,dx=\frac{(-1)^{p}{\left(1-\frac{1}{2}z\right)_{p}}}{2{\left(\frac{1}{2}+\frac{1}{2}z\right)_{p+1}}} \end{equation} where $(.)_n$ is Pochhammer's symbol, to express \begin{align} w_{2p+1}&=\int_0^1(1-x^2)P_{2p+1}(x)\,dx\\ &=\frac{(-1)^{p}{\left(\frac{1}{2}\right)_{p}}}{2{\left(1\right)_{p+1}}}-\frac{(-1)^{p}{\left(-\frac{1}{2}\right)_{p}}}{2{\left(2\right)_{p+1}}} \end{align} Then, converting the series into generalized hypergeometric functions (GHF), \begin{align} \varphi(R)&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)!}R^{n+1}w_n\\ &=w_0R+\frac16w_2R^3-\sum_{p=0}^\infty\frac{1}{(2p+2)!}w_{2p+1}R^{2p+2}\\ &=\frac23R-\frac1{45}R^3-\frac12\sum_{p=0}^\infty\frac{(-1)^p}{(2p+2)!}R^{2p+2}\left[\frac{\left(\frac{1}{2}\right)_{p}}{{\left(1\right)_{p+1}}}-\frac{\left(-\frac{1}{2}\right)_{p}}{{\left(2\right)_{p+1}}}\right]\\ &=\frac23R-\frac1{45}R^3-\frac{R^2}4\left[ \phantom{}_{2}F_{3}\! \left(\frac{1}{2},1;\frac{3}{2},2,2;-\frac{R^{2}}{4}\right) -\frac12\phantom{}_{2}F_{3}\! \left(-\frac{1}{2},1;\frac{3}{2},2,3;-\frac{R^{2}}{4}\right) \right] \end{align} Then the asymptotic behavior follows: \begin{equation} \varphi(R)=\frac23+\sqrt{\frac{2}{\pi}}\frac{\sin(R-\pi/4)}{R^{3/2}}-\frac{21}{8}\sqrt{\frac{2}{\pi}}\frac{\sin(R+\pi/4)}{R^{5/2}}+O(R^{-7/2}) \end{equation} This result was obtained with a CAS, however it should not be too difficult to derive it from the asymptotic expansion of the GHF.
This method should also apply to evaluate the second integral, by differentiating the starting identity \begin{equation} J_0(t\sin\theta)=e^{t\cos(\theta)}\sum_{n\ge0}\frac{(-1)^n}{n!}P_n(\cos\theta)t^n \end{equation} w.r.t. $t$ or $\theta$ to obtain an expansion of $J_1$, but I didn't try it.