I would like to integrate this in my research:
$\int_0^\infty s e^{i bs^2}J_0(a s)$, where a and b are both real and greater than zero. Integration by parts seems like the obvious first step, but that leaves a term like $\int_0^\infty e^{i bs^2}J_1(a s)$ , which seems even more complicated.
The topic is turbulence, and you can determine the answer on Mathematica for the definite integral (but I'd like to do it by hand).
Consider the more general integral \begin{align} I_{n} = \int_{0}^{\infty} t \, e^{-b t^{2}} \, J_{n}(a t) \, dt \end{align} for which \begin{align} I_{n} &= \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \left(\frac{a}{2}\right)^{2k+n}}{k! \, \Gamma(k+n+1)} \cdot \int_{0}^{\infty} e^{- b t^{2}} \, t^{2k+n+1} \, dt \\ &= \frac{1}{2} \, \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \left(\frac{a}{2}\right)^{2k+n}}{k! \, \Gamma(k+n+1)} \, \frac{\Gamma\left(k + \frac{n+2}{2}\right)}{b^{k+n/2 +1}} \\ &= \frac{\Gamma\left(\frac{n+2}{2}\right) \, (a/2)^{n}}{2 \, \Gamma(n+1) \, b^{n/2+1}} \, {}_{1}F_{1}\left(\frac{n}{2}+1; n+1; - \frac{a^{2}}{4b} \right) \\ &= \frac{\Gamma\left(\frac{n+1}{2}\right) \, \left(\frac{a}{\sqrt{b}}\right)^{n}}{2 \sqrt{\pi} \, b} \, {}_{1}F_{1}\left(\frac{n}{2}+1; n+1; - \frac{a^{2}}{4b} \right) \end{align} or \begin{align} \int_{0}^{\infty} t \, e^{-b t^{2}} \, J_{n}(a t) \, dt = \frac{\Gamma\left(\frac{n+1}{2}\right) \, \left(\frac{a}{\sqrt{b}}\right)^{n}}{2 \sqrt{\pi} \, b} \, {}_{1}F_{1}\left(\frac{n}{2}+1; n+1; - \frac{a^{2}}{4b} \right) \end{align}
When $n=0$ the reduction is \begin{align} \int_{0}^{\infty} t \, e^{-b t^{2}} \, J_{0}(a t) \, dt = \frac{1}{2 b} \, {}_{1}F_{1}\left(1; 1; - \frac{a^{2}}{4b} \right) = \frac{1}{2b} \, e^{-\frac{a^{2}}{4b}} \end{align} Now let $b \to -ib$ to obtain \begin{align} \int_{0}^{\infty} t \, e^{ib t^{2}} \, J_{0}(a t) \, dt = \frac{i}{2 b} \, e^{i\frac{a^{2}}{4b}} \end{align}
Hint: $$\mathcal{L}\left(J_0(a\sqrt{x})\right) = \frac{1}{t} e^{-\frac{a^2}{4t}},\tag{1}$$
$$\mathcal{L}\left(J_1(a\sqrt{x})\right) = \frac{|a|\sqrt{\pi}}{t^{3/2}}\left(I_0\left(e^{-\frac{a^2}{8t}}\right)-I_1\left(e^{-\frac{a^2}{8t}}\right)\right) e^{-\frac{a^2}{8t}}\tag{2}$$
$(1)$ just follows from writing $J_0$ as its Taylor series. The same technique applies for proving $(2)$.
So use the substitution $s=\sqrt{x}$.
Let $$ f(a,b)=\int_0^{+\infty}e^{ibs^2}J_0(as)\,s\,ds. $$ Changing variable to $t=\sqrt{b}s$ gives $f(a,b)=g(a/\sqrt{b})/b$, where $$ g(c)=\int_0^{+\infty}e^{it^2}tJ_0(ct)\,dt. $$ Integrating by parts, $$ \begin{aligned} g(c)&=\bigl[e^{it^2}/(2i) J_0(ct)\bigr]_0^{+\infty}-\int_0^{+\infty} e^{it^2}/(2i)(-c J_1(ct))\,dt\\ &=-\frac{1}{2i}+\frac{1}{2i}\int_0^{+\infty}e^{it^2}cJ_1(ct)\,dt. \end{aligned} $$ Using the fact that $cJ_1(ct)=\int_0^c \hat{c}tJ_0(\hat{c}t)\,d\hat{c}$, we get $$ g(c)=-\frac{1}{2i}+\frac{1}{2i}\int_0^c \hat{c}g(\hat{c})\,d\hat{c}. $$ Now $g$ is differentiable, and $$ g'(c)=\frac{1}{2i}cg(c),\quad g(0)=-\frac{1}{2i}. $$ This differential equation has solution $$ g(c)=-\frac{1}{2i}e^{-ic^2/4}, $$ and so $$ f(a,b)=\frac{g(a/\sqrt{b})}{b}=-\frac{1}{2ib}e^{-ia^2/(4b)}. $$
Comment
It is funny that, when differentiating $g$ under the integral sign with respect to $c$, we end up with the divergent integral $$ -\int_0^{+\infty}e^{it^2}t^2 J_1(ct)\,dt, $$ so one has to be a bit careful.
