The integral comes from a 2-D Fourier transform. $$ \int_{0}^{\theta_0} \frac{\sin(\theta)\cos(\theta)}{1+\cos(\theta)}J_{0}(a\sin(\theta))\mathrm{d}\theta $$ where $$ a\in\mathbb R^+, \theta_0\in(0,\pi) $$
I have drawn the function through numerical calculation as follows: enter image description here
The parameters is set as follows: $$ a=[0,9.425] , θ_0=\frac{\pi}{4}$$ So the $x$-axis represents $a$, $y$-axis represents the integration result.
I want to obtain an analytical solution but I do really bad at solving the complex integration.
So can someone point a possible way to solve it?
The standard Taylor expansion of the Bessel Function is $$ J_0(a\sin\theta) = 1 -\frac{1}{4} a^2\sin^2\theta +\frac{1}{64} a^4\sin^4\theta -\frac{1}{2304} a^6\sin^6\theta +\cdots ; $$ Interchange of the order of summation and integration yiels $$ \int_0^{\theta_0} \frac{\sin \theta \cos \theta}{1+\cos\theta}J_0(a\sin \theta)d\theta = \int_0^{\theta_0} \frac{\sin \theta \cos \theta}{1+\cos\theta}[ 1 -\frac{1}{4} a^2\sin^2\theta +\frac{1}{64} a^4\sin^4\theta -\frac{1}{2304} a^6\sin^6\theta +\frac{1}{147456} a^8\sin^8\theta +\cdots] d\theta $$ $$ = -\int_1^{\cos\theta_0} \frac{x}{1+x}[ 1 -\frac{1}{4} a^2(1-x^2) +\frac{1}{64} a^4(1-x^2)^2 -\frac{1}{2304} a^6(1-x^2)^3 +\cdots]dx $$ $$ = \int_{\cos\theta_0}^1 x[ 1 -\frac{1}{4} a^2(1-x) +\frac{1}{64} a^4(1-x^2)(1-x) -\frac{1}{2304} a^6(1-x^2)^2(1-x) +\cdots]dx $$ and all these intermediate polynomials have trivial underivatives: $$ \int x(1-x)(1-x^2)^k dx = \sum_{l=0}^k \binom{k}{l}(-1)^l\int x^{2l} x(1-x) dx $$ $$ = \sum_{l=0}^k \binom{k}{l}(-1)^l\int x^{2l+1} dx -\sum_{l=0}^k \binom{k}{l}(-1)^l\int x^{2l+2} dx $$ $$ = \sum_{l=0}^k \binom{k}{l}\frac{(-)^l}{2l+2} x^{2l+2} -\sum_{l=0}^k \binom{k}{l} \frac{(-)^l}{2l+3} x^{2l+3} $$