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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).

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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}

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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 $s=\sqrt{x}$.

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  • $\begingroup$ The result seems to be $i\exp(-ia^2/(4b))/(2b)$, do you get that from your calulation as well? $\endgroup$ – mickep Jun 25 '15 at 20:37
  • $\begingroup$ @mickep: yes, I agree. $\endgroup$ – Jack D'Aurizio Jun 25 '15 at 20:42
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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.

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