Thank you for reading my question!
As we know, the first kind Bessel function of 0th order can be integrated as follows $$ J_0(x) =\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-ix\sin{(a+\tau)}}d\tau =\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-ix\sin{\tau}}d\tau, a\in R $$ The constant $a$ will not change the value. The reason is explained by @GEdga in the comment.
However, this is the case for the following integration $$ \int_{0}^{\pi}e^{-ix\sin(\tau+a)}d\tau $$ Setting $x=1$, by the matlab code following, the absolute value would change with different $a$
int_func = @(x) exp(-1j*1*sin(x));
int_result = zeros(1,100);
x = linspace(1,100)/100*pi*2;
for i = 1:100
disp(x(i));
int_result(i) = integral(int_func,0+x(i),pi+x(i));
end
figure(1);
plot(x,abs(int_result),'LineWidth',1);
grid on;
xlabel('a');
ylabel('Integration result (abs)');
How can I get the analytical formula for this integration?
Thank you! Any help is appreciated.
Too long for a comment.
$$I=\int_{0}^{\pi}e^{-ix\sin(\tau+a)}\,d\tau$$
I faced the same integral quite recently but it was for small values of $a$ and $x$.
To solve the problem, I used a series expansion around $a=0$ and obtained
$$I=\pi (J_0(x)-i \pmb{H}_0(x))+ix \sum_{n=1}^\infty (-1)^{\frac{n+2}{2}}\,P_n(x)\,a^{2n}$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(x) \\ 1 & 1 \\ 2 & \frac{1}{12} \left(x^2+1\right) \\ 3 & \frac{1}{360} \left(x^4+10 x^2+1\right) \\ 4 & \frac{1}{20160} \left(x^6+35 x^4+91 x^2+1\right) \\ 5 & \frac{1}{1814400} \left(x^8+84 x^6+966 x^4+820 x^2+1\right) \\ 6 & \frac{1}{239500800} \left(x^{10}+165 x^8+5082 x^6+24970 x^4+7381 x^2+1\right) \\ \end{array} \right)$$
For sure, I took into account the fact that $\Re(I)=\pi J_0(x)$ does not depend on $a$.