A Sine integral: problem I

Question

Is it possible to demonstrate a solution for the integral \begin{align} \int_{0}^{\infty} x^{n} \, \sin\left( a x^{2} + \frac{b}{x^{2}} \right) \, dx \end{align}

Answer 1

Part 1

Lets assume that $n$ is even ($n=2m$), and the parameters are chosen that the integral converges. Call our integral of interest $J(a,b)$. Now observe that

\begin{align}x^{2m}\sin(ax^2+b/x^2)&= (-1)^{(m+3)/2}\frac{d^m}{da^m}\cos(ax^2+b/x^2),\qquad \text{if} \; m=1,3,5,\ldots \\ x^{2m}\sin(ax^2+b/x^2)&= (-1)^{m/2}\frac{d^m}{da^m}\sin(ax^2+b/x^2),\qquad\;\,\quad \text{if}\;m=2,4,6... \end{align}

So essentially we only have to calculate $\displaystyle\int_0^{\infty}\cos(ax^2+b/x^2)\mathrm{d}x$ and $\displaystyle\int_0^{\infty}\sin(ax^2+b/x^2)\mathrm{d}x$.

This can be done as follows (we only do the sine integral the other cosine one goes along the same lines):

1. First complete the square in the Integrands \begin{align} a/x^2+b/x^2= (\sqrt{a}x-\sqrt{b}/x)^2+2\sqrt{ab}, \end{align}

2. Use the trigonometric identity \begin{align} \sin(x+y)=\sin x\cos y+\cos x\sin y, \end{align} to obtain

\begin{align} _{e}I(a,b)&=\int_0^{\infty}\sin(ax^2+b/x^2)\mathrm{d}x \\ &= \cos(2\sqrt{ab})\int_0^{\infty}\sin(\sqrt{a}x-\sqrt{b}/x)^2\mathrm{d}x+\sin(2\sqrt{ab})\int_0^{\infty}\cos(\sqrt{a}x-\sqrt{b}/x)^2\mathrm{d}x \end{align}

3. Now we apply the inversion $\displaystyle x \rightarrow \frac{\sqrt{b}}{\sqrt{a}x}$ to both integrals and add them together with their non-inverted versions. We end up with

\begin{align} 2 _{e}I(a,b) &= \cos(2\sqrt{ab})\int_0^{\infty}\sin(\sqrt{a}x-\sqrt{b}/x)^2 \left(1+\frac{\sqrt{b}}{\sqrt{a}x^2}\right)\mathrm{d}x \\ & \hspace{10mm} +\sin(2\sqrt{ab})\int_0^{\infty}\cos(\sqrt{a}x-\sqrt{b}/x)^2\left(1+\frac{\sqrt{b}}{\sqrt{a}x^2}\right)\mathrm{d}x. \end{align}

4. Substitute $\sqrt{a}x-\sqrt{b}/x=y$ and $\displaystyle\frac{dy}{\sqrt{a}}=\left(1+\frac{\sqrt{b}}{\sqrt{a} x^2}\right)\mathrm{d}x$ to get

\begin{align} 2 _{e}I(a,b) &=\frac{\cos(2\sqrt{ab})}{\sqrt{a}} \int_{-\infty}^{\infty} \sin y^2\mathrm{d}y + \frac{\sin(2\sqrt{ab})}{\sqrt{a}} \int_{-\infty}^{\infty}\cos y^2\mathrm{d}y \\ &= \frac{2 \, \cos(2\sqrt{ab})}{\sqrt{a}} \int_{0}^{\infty} \sin y^2\mathrm{d}y + \frac{2\, \sin(2\sqrt{ab})}{\sqrt{a}} \int_{0}^{\infty}\cos y^2 \mathrm{d}y. \end{align}

5. Use the well known values $\displaystyle\int_0^{\infty}\sin y^2\mathrm{d}y=\int_0^{\infty}\cos y^2\mathrm{d}y=\sqrt{\frac{\pi}{8}}$.

So in the end we have

\begin{align} _{e}I(a,b) = \sqrt{\frac{\pi}{8 a}}\left(\cos(2\sqrt{ab})+\sin(2\sqrt{ab})\right) = \frac{1}{2} \sqrt{\frac{\pi}{a}} \, \sin\left( 2 \sqrt{ab} + \frac{\pi}{4} \right). \end{align}

The final result is now

\begin{align} J(a,b)=\int_0^{\infty}x^{2m}\sin(ax^2+b/x^2)\mathrm{d}x=(-1)^{(m+3)/2}\frac{d^m}{da^m}_{e}I(a,b), \end{align} if $m=1,3,5\ldots$

Edit: For $m=2,4,6\ldots$ we get

\begin{align} _{e}K(a,b) &= \int_0^{\infty}\cos(ax^2+b/x^2)\mathrm{d}x = \sqrt{\frac{\pi}{8 a}}\left(\cos(2\sqrt{ab})-\sin(2\sqrt{ab})\right) \\ &= \frac{1}{2} \sqrt{ \frac{\pi}{a} } \, \cos\left( 2 \sqrt{ab} + \frac{\pi}{4} \right), \end{align} where we used $\cos(x+y)=\cos x\cos y-\sin x \sin y$ and \begin{align} J(a,b)=\int_0^{\infty}x^{2m}\cos(ax^2+b/x^2)\mathrm{d}x=(-1)^{m/2}\frac{d^m}{da^m}{_{e}K(a,b)}. \end{align}

Part 2

I finally managed managed to get an answer for $n=2m+1$ We again observe the differentiation trick works again:

\begin{align}x^{2m+1}\sin(ax^2+b/x^2)&= (-1)^{(m+3)/2}\frac{d^m}{da^m}x\cos(ax^2+b/x^2), \qquad \text{if}\; m=1,3,5\ldots \\ x^{2m+1}\sin(ax^2+b/x^2)&= (-1)^{m/2}\frac{d^m}{da^m}x \sin(ax^2+b/x^2),\qquad\quad\;\, \text{if} \; m=2,4,6\ldots \end{align} So we only need $\displaystyle\int_0^{\infty}x\cos(ax^2+b/x^2)\mathrm{d}x$ and $\displaystyle\int_0^{\infty}x\sin(ax^2+b/x^2)\mathrm{d}x$.

We do this as follows (only the sine case in detail...)

1. Substitute $x\rightarrow\sqrt{y}$,$dx=dy/\sqrt{y}$ and luckily the intergal simplifies a lot: \begin{align} _oI(a,b)=\int_0^{\infty}\sin(ay+b/y)\mathrm{d}y. \end{align}

2. Observe that our Integral can after rescaling $y=\sqrt\frac{b}{a}y$ be written as \begin{align} _oI(a,b)=-\sqrt{\frac{b}{a}}\Im\int_0^{\infty}e^{-\sqrt{ab}i(y+1/y)}\mathrm{d}y \end{align}

3. Referring to the the beautiful answers in this link and analytically (which is as far as I can see legitimate) continuing them to purely imaginary parameters ($a-\rightarrow ia$,$b\rightarrow ib$) we can write \begin{align} _oI(a,b)=-2\sqrt{\frac{b}{a}}\Im K_1(2i\sqrt{ab}) \end{align} where $K_1(z)$ is a modified Bessel function

4. Using furthermore the Relations $K_1(z)=\frac{i \pi}{2}e^{i \pi/2}H_1(iz)$ and $H_1(z)=J_1(z)+iN_1(z)$, with $H_1(z)$ an Hankel function and $J_1(z)$/$N_1(z)$ Bessel functions of first/second kind, see here for example, we finally get

\begin{align} _oI(a,b)=-\pi\sqrt{\frac{b}{a}}N_1(2\sqrt{ab}) \end{align}

5. Repeating this procedure (with $\Re$ instead of $\Im$) for the Cosine case we get

\begin{align} _oK(a,b)=-\pi\sqrt{\frac{b}{a}}J_1(2\sqrt{ab}). \end{align}

The whole Integral is then given by

\begin{align} J(a,b)=(-1)^{(m+3)/2}\frac{d^m}{da^m}{_{o}I(a,b)}, \end{align} if $m=1,3,5\ldots$ or \begin{align} J(a,b)=(-1)^{m/2}\frac{d^m}{da^m}{_{o}K(a,b)}, \end{align} if $m=2,4,6\ldots$

Answer 2

Case $1$: $a>0$ and $b>0$

Then $\int_0^\infty x^n\sin\left(ax^2+\dfrac{b}{x^2}\right)dx$

$=\int_0^\infty x^\frac{n}{2}\sin\left(ax+\dfrac{b}{x}\right)d\left(x^\frac{1}{2}\right)$

$=\dfrac{1}{2}\int_0^\infty x^\frac{n-1}{2}\sin\left(ax+\dfrac{b}{x}\right)dx$

$=\dfrac{1}{2}\int_0^\infty\left(\dfrac{\sqrt bx}{\sqrt a}\right)^\frac{n-1}{2}\sin\left(a\dfrac{\sqrt bx}{\sqrt a}+\dfrac{b}{\dfrac{\sqrt bx}{\sqrt a}}\right)d\left(\dfrac{\sqrt bx}{\sqrt a}\right)$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_0^\infty x^\frac{n-1}{2}\sin\left(\sqrt{ab}\left(x+\dfrac{1}{x}\right)\right)dx$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_{-\infty}^\infty(e^x)^\frac{n-1}{2}\sin\left(\sqrt{ab}\left(e^x+\dfrac{1}{e^x}\right)\right)d(e^x)$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_{-\infty}^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(\int_{-\infty}^0e^\frac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx\right)$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(\int_\infty^0e^\frac{(n+1)(-x)}{2}\sin\left(2\sqrt{ab}\cosh(-x)\right)d(-x)+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx\right)$

$=\dfrac{b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(\int_0^\infty e^{-\frac{(n+1)x}{2}}\sin\left(2\sqrt{ab}\cosh x\right)dx+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx\right)$

$=\dfrac{b^\frac{n+1}{4}}{a^\frac{n+1}{4}}\int_0^\infty\cosh\dfrac{(n+1)x}{2}\sin\left(2\sqrt{ab}\cosh x\right)dx$

$=\dfrac{\pi b^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(J_\frac{n+1}{2}\left(2\sqrt{ab}\right)\cos\dfrac{(n+1)\pi}{4}-Y_\frac{n+1}{2}\left(2\sqrt{ab}\right)\sin\dfrac{(n+1)\pi}{4}\right)$ (according to http://dlmf.nist.gov/10.9#E8)

Case $2$: $a>0$ and $b<0$

Then $\int_0^\infty x^n\sin\left(ax^2+\dfrac{b}{x^2}\right)dx$

$=\int_0^\infty x^\frac{n}{2}\sin\left(ax+\dfrac{b}{x}\right)d\left(x^\frac{1}{2}\right)$

$=\dfrac{1}{2}\int_0^\infty x^\frac{n-1}{2}\sin\left(ax+\dfrac{b}{x}\right)dx$

$=\dfrac{1}{2}\int_0^\infty\left(\dfrac{\sqrt{-b}x}{\sqrt a}\right)^\frac{n-1}{2}\sin\left(a\dfrac{\sqrt{-b}x}{\sqrt a}+\dfrac{b}{\dfrac{\sqrt{-b}x}{\sqrt a}}\right)d\left(\dfrac{\sqrt{-b}x}{\sqrt a}\right)$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_0^\infty x^\frac{n-1}{2}\sin\left(\sqrt{-ab}\left(x-\dfrac{1}{x}\right)\right)dx$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_{-\infty}^\infty(e^x)^\frac{n-1}{2}\sin\left(\sqrt{-ab}\left(e^x-\dfrac{1}{e^x}\right)\right)d(e^x)$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\int_{-\infty}^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(\int_{-\infty}^0e^\frac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx\right)$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(\int_\infty^0e^\frac{(n+1)(-x)}{2}\sin\left(2\sqrt{-ab}\sinh(-x)\right)d(-x)+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx\right)$

$=\dfrac{(-b)^\frac{n+1}{4}}{2a^\frac{n+1}{4}}\left(-\int_0^\infty e^{-\frac{(n+1)x}{2}}\sin\left(2\sqrt{-ab}\sinh x\right)dx+\int_0^\infty e^\frac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx\right)$

$=\dfrac{(-b)^\frac{n+1}{4}}{a^\frac{n+1}{4}}\int_0^\infty\sinh\dfrac{(n+1)x}{2}\sin\left(2\sqrt{-ab}\sinh x\right)dx$

$=\dfrac{(-b)^\frac{n+1}{4}}{a^\frac{n+1}{4}}K_\frac{n+1}{2}\left(2\sqrt{-ab}\right)\sin\dfrac{(n+1)\pi}{4}$ (according to http://people.math.sfu.ca/~cbm/aands/page_376.htm)

Case $3$: $a<0$ and $b>0$

Then $\int_0^\infty x^n\sin\left(ax^2+\dfrac{b}{x^2}\right)dx=-\int_0^\infty x^n\sin\left(-ax^2-\dfrac{b}{x^2}\right)dx$

Follow the Case $2$

Case $4$: $a<0$ and $b<0$

Then $\int_0^\infty x^n\sin\left(ax^2+\dfrac{b}{x^2}\right)dx=-\int_0^\infty x^n\sin\left(-ax^2-\dfrac{b}{x^2}\right)dx$

Follow the Case $1$