Can this integral be calculated in closed form?

Solution 1:

This is not an answer either, but MSE wouldn't let me edit my comment.

Let $M(a,b)$ be your integral, and let $N(a,b)$ be the integral you found in the table with $x^3$. Then $M$ and $N$ are related by the ODE: $M(a,b)-N(a,b)=\frac{\partial^2 M}{\partial a^2}(a,b)$, which can be solved for $M(a,b)$ in terms of $N$ (which is in your table), $M(0,b)=0$, and $\frac{\partial M}{\partial a}(0,b)$ in the usual way. According to WolframAlpha, $$\frac{\partial M}{\partial a}(0,b)=\int_0^1xJ_2(bx)\sqrt{1-x^2}dx=\frac{b(2+\cos b)-3\sin b}{b^3}.$$ So if $N(a,b)$ is sufficiently nice, you might have a chance!

Solution 2:

This is not an answer but it is too long for a comment.

Inspired by Jack D'Aurizio's first comment, let us write the series

$$J_2 (b x)=\sum_{m=0}^\infty \frac{(-1)^m}{m!(m+2)!} \left(\frac {b}2\right)^{2m+2}x^{2m+2}$$ $$\sin(a\sqrt{1-x^2})=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}a^{2n+1}\left(1-x^2\right)^{n+\frac 12}$$ which let us with a double summation of integrals $$I_{m,n}=\int_0^1 x^{2m+3}\left(1-x^2\right)^{n+\frac 12}\,dx=\frac{\Gamma (m+2) \Gamma \left(n+\frac{3}{2}\right)}{2\, \Gamma \left(m+n+\frac{7}{2}\right)}$$ which leads to

$$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=\sqrt{\pi }\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } (-1)^{m+n}\frac{ a^{2 n+1}\, b^{2 m+2} }{2^{2 (m+n+2)}(m+2) \,m! \,n!\, \Gamma \left(m+n+\frac{7}{2}\right)}$$ which, looking at some numerical tests, seems to converge quite fast.

For illustration purposes, using $\sum _{m=0}^{p } \sum _{n=0}^{p }$, $a=3$ and $b=5$ the following decimal representations are obtained $$\left( \begin{array}{cc} p & \text{result} \\ 2 & 0.263943 \\ 3 & 0.083938 \\ 4 & 0.119953 \\ 5 & 0.114727 \\ 6 & 0.115299 \\ 7 & 0.115250 \\ 8 & 0.115253 \end{array} \right)$$ which is identical (at least for six significant figures) to the result of the numerical integration.

Edit

May be interesting is that, thanks to a CAS, summing over $n$, the integral can be expressed as $$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=\sqrt{\pi }\sum _{m=0}^{\infty }(-1)^m\frac{b^{2m+2} }{(2a)^{m+\frac 32}\, (m+2)\, m! }\,J_{m+\frac{5}{2}}(a)$$

$$\int^1_0 x J_2 (b x) \sin(a\sqrt{1-x^2})\, dx=2a\sum _{m=0}^{\infty }(-1)^m\frac{ (m+1) b^{2 m+2} }{\Gamma (2 m+6)}\, _0F_1\left(;m+\frac{7}{2};-\frac{a^2}{4}\right)$$

Using the same example as above $$\left( \begin{array}{cc} m & \sum_{k=0}^m \\ 2 & 0.249952 \\ 3 & 0.085208 \\ 4 & 0.119870 \\ 5 & 0.114731 \\ 6 & 0.115298 \\ 7 & 0.115250 \\ 8 & 0.115253 \end{array} \right)$$