Approximation of $ \int_0^\infty x^{1/2} e^{-x t} \sin (a x^{1/2})dx$ -- rapidly oscillating

Question

The exact solution of the integral $$ I(t) = \int_0^\infty x^{1/2} e^{-x t} \sin (a x^{1/2})dx$$ according to mathematica is $$ \frac{a \sqrt{t}-\left(a^2-2 t\right) F\left(\frac{a}{2 \sqrt{t}}\right)}{2 t^{5/2}}.$$ At large $t$ this result is dominated by the slowest-to-decay term: $I(t) \sim t^{-2}$.

Is there a means by which I can derive this asymptotic result directly? Is it also possible to access the $t \rightarrow 0$ form approximately? This is a rapidly oscillating integral, so I believe the method of stationary phase should be applicable. I know the critical points of the oscillating term are at $$ x = \Big(\frac{\pi}{2a}\Big)^2(2n+1)^2,$$ where $n$ is a positive integer, so I guess I can write the integrand as a sum of the integrand expanded around these critical points, as in something like $$ I(t) \approx \sum_{n=0}^\infty (-)^n \int_0^\infty dx \exp[a + b(x-x_n) + c (x-x_n)^2], $$ but I am unsure how to achieve this. Can anyone offer some guidance?

Answer 1

By the power series expansion of the sine function, $$ x^{1/2} \sin (ax^{1/2} ) = \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{a^{2n + 1} }}{{(2n + 1)!}}x^{n + 1} } $$ for any $x\geq 0$. Substitution into the integral and integrating term-by-term gives $$ I(t) = \frac{a}{{t^2 }}\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{(n + 1)!}}{{(2n + 1)!}}\left( {\frac{{a^2 }}{t}} \right)^n } . $$ This series converges for any $a\in \mathbb{R}$ and $t>0$ since $$ \frac{{(n + 1)!}}{{(2n + 1)!}} \sim \frac{{\sqrt {\pi n} }}{{2^{2n + 1} n!}},\quad n\to +\infty. $$ To justify the term-wise integration, one can use the fact that the absolute value of the remainder in the series for the sine is at most the absolute value of the first omitted term.

Note that the series in fact converges for all $t\in \mathbb C \setminus \left\{0\right\}$ and $a \in \mathbb{C}$, and $I(t)$ has an essential singularity at $t=0$ (if $a\neq 0$).

Addendum. Elaboration on what I said about the justification of interchanging the order of summation and integration. Using the known properties of the Maclaurin series of the sine, we have, for any $N\geq 0$, $a\in \mathbb{R}$ and $x>0$, $$ x^{1/2} \sin (ax^{1/2} ) = \sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{a^{2n + 1} }}{{(2n + 1)!}}x^{n + 1} } + ( - 1)^N \frac{{a^{2N + 1} }}{{(2N + 1)!}}x^{N + 1} \theta _N (x,a) $$ with a suitable $0 < \theta _N (x,a) < 1$. Thus, employing the mean value theorem for improper integrals, we find, for any $N\geq 0$, $a\in \mathbb{R}$ and $t>0$, $$ I(t) = \frac{a}{{t^2 }}\left( {\sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{{(n + 1)!}}{{(2n + 1)!}}\left( {\frac{{a^2 }}{t}} \right)^n } + ( - 1)^N \frac{{(N + 1)!}}{{(2N + 1)!}}\left( {\frac{{a^2 }}{t}} \right)^N \Theta _N (t,a)} \right) $$ with an appropriate $0 < \Theta _N (t,a) < 1$. Letting $N\to +\infty$ and applying the above estimate for the ratio of factorials yields the desired series expansion.

Answer 2

$\newcommand{\d}{\,\mathrm{d}}$I will assume $a,t,$ are real constants.$$\sin(ax^{1/2})=\Im\exp(i\cdot ax^{1/2})$$

Then consider:

$$\Im\int_0^\infty x^{1/2}\exp(-xt)\exp(i\cdot ax^{1/2})\d x=\Im\int_0^\infty x^{1/2}\exp\left(-t\left(x^{1/2}-i\cdot\frac{a}{2t}\right)^2-\frac{a^2}{4t}\right)\d x$$

We substitute $x\mapsto x^2$, $\d x\mapsto 2x\d x$, and then shift the variable:

$$2\exp\left(-\frac{a^2}{4t}\right)\cdot\Im\int_0^\infty x^2\exp\left(-t\left(x-i\cdot\frac{a}{2t}\right)^2\right)\d x=\\\hspace{30pt}2\exp\left(-\frac{a^2}{4t}\right)\cdot\Im\int_{-ia/2t}^{\infty-ia/2t}\left(x^2+i\cdot\frac{a}{t}x-\frac{a^2}{4t^2}\right)\exp(-tx^2)\d x\\=2\exp\left(-\frac{a^2}{4t}\right)\cdot\Im\left[\int_{-ia/2t}^{\infty-ia/2t} x^2\exp(-tx^2)\d x\\\hspace{35pt}+\color{red}{i\cdot\frac{a}{t}\int_{-ia/2t}^{\infty-ia/2t}x\exp(-tx^2)\d x}\\\hspace{30pt}-\frac{a^2}{4t^2}\int_{-ia/2t}^{\infty-ia/2t}\exp(-tx^2)\d x\right]\\=\color{red}{2\frac{a}{t}\exp\left(-\frac{a^2}{4t}\right)\Re\left[-\frac{\exp(-tx^2)}{2t}\right]_{-ia/2t}^{\infty-ia/2t}}\\\hspace{80pt}+2\exp\left(-\frac{a^2}{4t}\right)\Im\left[\int_{-ia/2t}^{\infty-ia/2t}\left(x^2-\frac{a^2}{4t}\right)\exp(-tx^2)\d x\right]\\=2\frac{a}{t}\exp\left(-\frac{a^2}{4t}\right)\Re\left[\frac{\exp(-t(-ia/2t)^2)}{2t}\right]\\\hspace{30pt}+2\exp\left(-\frac{a^2}{4t}\right)\Im\left[\int_{0}^{\infty}\left(x^2-i\cdot\frac{a}{t}x-\frac{a^2}{2t^2}\right)\exp\left(-tx^2+i\cdot ax+\frac{a^2}{4t}\right)\d x\right]\\=\frac{a}{t^2}-2\frac{a}{t}\int_0^\infty x\exp(-tx^2)\cos(ax)\d x+2\int_0^\infty x^2\sin(ax)\exp(-tx^2)\d x\\\hspace{30pt}-\frac{a^2}{t^2}\int_0^\infty\exp(-tx^2)\sin(ax)\d x$$

Let's evaluate the cosine integral by parts:

$$-2\frac{a}{t}\left[-\frac{\exp(-tx^2)}{2t}\cos(ax)\right]_0^\infty+\frac{a^2}{t^2}\int_0^\infty\exp(-tx^2)\sin(ax)\d x\\\hspace{50pt}=-\frac{a}{t^2}+\frac{a^2}{t^{5/2}}\int_0^\infty\exp(-x^2)\sin\left(\frac{a}{\sqrt{t}}x\right)\d x=-\frac{a}{t^2}+\frac{a^2}{t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)$$

Now for the middle integral, integrating by parts $x\exp(-tx^2)$ and differentiating $x\sin(ax)$:

$$2\left[-\frac{\exp(-tx^2)}{2t}x\sin(ax)\right]_0^\infty+\frac{1}{t}\int_0^\infty\exp(-tx^2)\sin(ax)\d x+\frac{a}{t}\int_0^\infty\exp(-tx^2)x\cos(ax)\d x$$

But we have already evaluated the latter two integrals, and the first term is zero, so we have:

$$\frac{1}{t^{3/2}}F\left(\frac{a}{2\sqrt{t}}\right)+\frac{a}{2t^2}-\frac{a^2}{2t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)$$

Using the same special function $F$ on the last integral, we find:

$$-\frac{a^2}{t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)$$

Summing all terms:

$$\begin{align}I(t)&=\frac{a^2}{t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)+\frac{1}{t^{3/2}}F\left(\frac{a}{2\sqrt{t}}\right)+\frac{a}{2t^2}-\frac{a^2}{2t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)-\frac{a^2}{t^{5/2}}F\left(\frac{a}{2\sqrt{t}}\right)\\&=\frac{a}{2t^2}+F\left(\frac{a}{2\sqrt{t}}\right)\left[\frac{1}{t^{3/2}}-\frac{a^2}{2t^{5/2}}\right]\end{align}$$

Which is the same as Mathematica’s answer, and more importantly a sketch on Desmos finds the numerical evaluation to be actually correct! Mathematica occasionally gives false answers so it is worth a check...

If you wanted to just see the asymptotics as $t\to\infty$, you could have noticed, without fully calculating, from where I highlighted in red earlier, the $\sim t^{-2}$ relation. The other terms were the imaginary parts of integrals very close to the real axis, which were also being multiplied by an $\exp(1/t^2)$ term, and thus quite negligible. You could expand the $\exp$ to get a more thorough analysis, but for the sake of asymptotes you could have stopped after evaluating the red term.

By the way, don’t worry if Dawson’s integral is new to you, because I myself have never seen this Dawson function $F$ before. I am just told that:

$$\int_0^\infty\exp(-x^2)\sin\left(\frac{a}{\sqrt{t}}x\right)\d x=F\left(\frac{a}{2\sqrt{t}}\right)$$

Which is all I need to know!

Answer 3

This is not an answer since it uses the explicit representation in terms of the Dawson function

$$I=\int_0^\infty \sqrt{x}\, e^{-t x}\, \sin \left(a \sqrt{x}\right)\,dx$$

Let $a \sqrt{x}=y$ and $k=\frac t{a^2}$ to make $$J=2a^3\,I=\frac 1{k^2}+\frac{2k-1}{k^{5/2}}\,F\left(\frac{1}{2 \sqrt{k}}\right) \tag 1$$ Expanded as a series for large values of $k$, we have $$J=\frac 2{k^2}\sum_{n=0}^\infty (-1)^n \frac{(n+1)! }{(2 n+1)!}k^{-n}\tag 2$$ and provided a sufficient large number of terms, this expansion is quite good from the maximum of $(1)$ to $\infty$ (the maximum occurs at $k=0.219873$).

Now, for very small values of $k$, we have $$J=-4\sum_{n=0}^\infty \frac{(2 n+2)! }{n!}k^{n}$$ which is quite good from $0$ to half the minimum of $(1)$ which corresponds to $k=0.045636$