A Sine integral: problem I

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}

Solutions Collecting From Web of "A Sine integral: problem I"

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

\text{if} \; m=1,3,5,\ldots
\\
\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}

1. 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}

1. 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}

1. 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),
\\
x^{2m+1}\sin(ax^2+b/x^2)&= (-1)^{m/2}\frac{d^m}{da^m}x
\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}

1. 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 $m=2,4,6\ldots$

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 $1$