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I was wondering if the following integral has a closed-form solution? $$I(x) = \int J_0(x)\sin(ax)\mathrm{d}x$$ where $a$ is a constant. I know the answer for the case when $a=1$, see here. I tried the similar method in that link but I was stuck. Integrating by parts yields $$ I(x) =x J_0(x)\sin(ax) - \int x [-J_1(x)\sin(ax) + aJ_0(x)\cos(ax)\mathrm{d}x $$

If $a=1$, we can continue by $$ \begin{split} I(x) &= xJ_0(x)\sin x - \int [xJ_1(x)\cos x]'\mathrm{d} x\\ &=xJ_0(x)\sin x -x J_1(x)\cos x \end{split} $$ where the relation $[xJ_1(x)]'=xJ_0(x)$ has been used. I have no idea for the case when $a\neq 1$.

Thanks in advance.

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    $\begingroup$ What about differentiating with respect to $a$? $\endgroup$ Oct 12, 2021 at 8:17
  • $\begingroup$ @blamethelag Could you please give more hints? I cannot follow. $\endgroup$ Oct 14, 2021 at 5:38
  • $\begingroup$ do you know about differentiation under the integral sign? $\endgroup$ Oct 14, 2021 at 9:19
  • $\begingroup$ @blamethelag. Could you go ahead ? I think that this makes the problem still more difficult ... but I may be wrong. Cheers :-) $\endgroup$ Oct 14, 2021 at 10:20

3 Answers 3

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I am skeptical about a possible closed form for the case where $a\neq 1$.

However, we could use the series expansion of $\sin(ax)$ and face the problem of $$\sum_{n=0}^\infty (-1)^n\frac{ a^{2 n+1} }{(2 n+1)!}\int x^{2 n+1} J_0(x)\,dx$$ and $$\int x^{2 n+1} J_0(x)\,dx=\frac{ \Gamma (n+1) }{2} x^{2 n+2}\, _1\tilde{F}_2\left(n+1;1,n+2;-\frac{x^2}{4}\right)$$ where appears the regularized generalized hypergeometric function.

Even if the first terms are quite large, the partial sums (from $n=0$ to $n=p$) converge quite fast.

Edit

We also could use the series expansion of $J_0(x)$ and face the problem of $$\sum_{n=0}^\infty \frac {(-1)^n } {4^n \, [n!]^2}\int x^{2n} \sin(ax)\,dx$$ $$\int x^{2n} \sin(ax)\,dx=\frac 1{a^{2n+1}}\int y^{2n}\, \sin(y)\,dy$$ $$\int y^{2n}\, \sin(y)\,dy=\Im\Big[(-i)^{n+1} \Gamma (2 n+1,-i y) \Big]$$

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  • $\begingroup$ For this approach, I know $\int x^{2n+1} J_0(x)\mathrm{d}x = -\sum_{m=0}^n \epsilon_m J_{2m}(x)$, where $\epsilon_m$ is the Neumann factor, i.e., $\epsilon_0 = 1$, and $\epsilon_m = 2$ when $m\neq 0$. This enables no requirement of the evaluation of the hypergeometric function. However, I expect a solution without an infinite series. $\endgroup$ Oct 14, 2021 at 5:53
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Note $I(a,x)$ your integral $$ \int_{-\infty}^x J_0(t)\sin(at)dt $$ and differentite twice under the integral sign: $$ \frac{d^2}{da^2}I(a,x)= \int_{-\infty}^x \frac{d^2}{da^2}J_0(t)\sin(at)dt = -a^2I(a,x). $$ So fix $x$, the function $f = I(\cdot,x)$ satisfies an ODE, you can find your function $f$ solving the ODE. The solution can be computed explicitely. Denote $X(a) = (f'(a), f(a))$, then you obtain the vector equation $$ X'(a) = A(a)X,~~A(a) = \left( \begin{array}{c c} 0 & -a^2 \\ 1 & 0 \\ \end{array} \right). $$

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    $\begingroup$ I'm not convinced about the result of differentiation: one gets a factor of $t^2$ in the integrand, which does not look like it reduces to simply $-a^2 I$ $\endgroup$
    – Sal
    Dec 3, 2021 at 1:38
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Let’s use a good solution from @Claude Leibovici using the series expansion for the Regularized Hypergeometric function:

$$\int \text J_0(x)\sin(ax)dx=\sum_{n=0}^\infty (-1)^n\frac{ a^{2 n+1} }{(2 n+1)!}\int x^{2 n+1} J_0(x)\,dx=C+\sum_{m=0}^\infty \frac{(-1)^m a^{2m+1}}{(2m+1)!}\sum_{n=0}^\infty \frac{(m+1)_n \left(-\frac{x^2}{4}\right)^n}{(1)_n (m+2)_n n!}$$

Simplifying the pochhammer symbols and using the Legendre Duplication formula

$$C+\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(-1)^m a^{2m+1}x^{2m+2} \left(-\frac{x^2}{4}\right)^n}{2\Gamma(2(m+1))(m+n+1)(1)_n n!}= C+\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{\sqrt \pi(-1)^m a^{2m+1}x^{2m+2} \left(-\frac{x^2}{4}\right)^n}{2^{2m+1}2\Gamma(m+1)\Gamma\left(m+\frac32\right)(m+n+1)(1)_n n!} =C+\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(1)_{m+n}\frac{ax^2}{2}\left(\frac{a^2x^2}{4}\right)^m\left(-\frac{x^2}4\right)^n}{(2)_{m+n}\left(\frac32\right)_m(1)_nm!n!}$$

Now let’s use the Kampé de Fériet function which also appears on Wolfram Functions defined as:

$$\text F^{p,r,u}_{q,s,v}\left(^{a_1,…,a_p;c_1,…,c_r;f_1,…,f_u}_{b_1,…,b_q;d_1,…,d_s;g_1,…,g_v}\ x,y\right)\mathop=^\text{def}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{\prod\limits_{j=1}^p(a_j)_{m+n} \prod\limits_{j=1}^r(c_j)_m \prod\limits_{j=1}^u (f_j)_n x^my^n}{\prod\limits_{j=1}^q (b_j)_{m+n} \prod\limits_{j=1}^s(d_j)_m \prod\limits_{j=1}^v(g_j)_n m!n!}$$

where there is the Pochhammer Symbol $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$ and therefore:

$$\int \text J_0(x)\sin(ax)dx= C+\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(1)_{m+n}\frac{ax^2}{2}\left(\frac{a^2x^2}{4}\right)^m\left(-\frac{x^2}4\right)^n}{(2)_{m+n}\left(\frac32\right)_m(1)_nm!n!}=\frac{ax^2}{2}\text F^{1,0,0}_{1,1,1}\left(^{\ \ 1;;}_{2;\frac32;1}\ \frac{a^2x^2}4,-\frac{x^2}4\right)+C $$

Which looks like an Appell or Horn Hypergeometric function, but none of these functions have the “$\text F^{1,0,0}_{1,1,1}$“, for example the The First Appell Hypergeometric function is defined in a similar way:

$$\text F_1(a;b_1,b_2;c;z_1,z_2)=\text F^{1,1,1}_{1,0,0}\left(^{a;b_1;b_2}_{\ \ \ c;;}\ z_1,z_2\right)$$

So we can have a closed form in terms of a general function which probably will simplify into other functions:

We have the $(2n+1)!$ in the denominator which will complicate, but using the Legendre Duplication theorem.The expansion for sine should have a large radius of convergence. Let’s see if there are any reduction formulas that we could use. Please correct me and give me feedback!

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  • $\begingroup$ If it were the case that $\partial_{aa}y=-a^2 y$, then after getting the solution in terms of the $D_{-1/2}$, the constants $c_1$ and $c_2$ could be found easily by evaluating $y$ at $a=0$ and $a=1$. Unfortunately it looks like this is not the case. Did you check the differentiation? $\endgroup$
    – Sal
    Dec 3, 2021 at 1:42
  • $\begingroup$ $\frac{d}{da}\sin(ax)=x \cos(ax) \neq a \cos(ax)$ $\endgroup$
    – Sal
    Dec 3, 2021 at 22:01
  • $\begingroup$ @Sal Thanks. I mixed up the chain rule there. Then the result should be: $$\frac{d^2}{da^2}\int \text J_0(x)\sin(ax)dx=-\int x^2 \text J_0(x)\sin(ax)dx$$ $\endgroup$ Dec 3, 2021 at 22:34
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    $\begingroup$ Correct. And unfortunately the new integral with $x^2$ is just as bad as what we started with! $\endgroup$
    – Sal
    Dec 3, 2021 at 22:40

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