$\int_{-\pi/2}^{\pi/2} \sin{\varphi} \ e^{jka\sin{\varphi}} \ d\varphi$ =?

Question

Could you help me solve this integral

\begin{equation} \int_{-\pi/2}^{\pi/2} \sin{\varphi} \ e^{jka\sin{\varphi}} \ d\varphi \end{equation}

Thank you so much!

Answer 1

Cylindrical Bessel function $J_0(z)$ is defined as $$J_0(z)=\pi^{-1}\int_{0} ^{\pi} \cos(z \cos x) dx ~~\mbox{and}~~ J'_0(z)=-J_1(z) = - \pi^{-1}\int_0^{\pi} \cos x~ \sin(z \cos x)~ dx~~(1). $$Denote the given integral by $I$. So, $$I=\int_{-\pi/2}^{\pi/2} \sin x ~ e^{iz\sin x} dx= \int_{0}^{pi/2} \sin x ~[e^{iz \sin x} -e^{-iz\sin x}] dx=2i \int_{0}^{\pi/2} \sin x ~\sin(z \sin x)~dx.$$ Next use $$\int_{0}^{a} f(x)~dx = \int_{0}^{a} f(a-x) ~dx$$ To get $$I=2 i \int_{0}^{\pi/2} \cos x ~\sin (z ~\cos x)~ dx.$$ Now use the symmetry property $$2\int_{0}^{a} f(x)~ dx= \int_{0}^{2a} f(x) ~dx, ~~\mbox{if}~~ f(2a-x)=f(x)$$ Then $$I=i\int_{0}^{\pi} \cos x \sin(z \cos x) dx =i \pi J_1(z)~~\mbox{see (1).} $$ Hence the answer is $i\pi J_1(z).$