Asymptotic behavior of the integral $H(\beta)=\frac{2}{\pi}\beta\int_{0}^{\infty}\exp\left(-x^{3/2}\right)\sin(\beta x)xdx$

Question

I found the integral $H(\beta)$ (which is called Holtsmark distribution) in Holtsmark's theory of ion field in plasma. In a book there is its asymptotic representation at small and great $\beta$:

$$ H(\beta)\approx \left\{\begin{array}{l} \frac{4\beta^2}{3\pi}\left(1-0,463\beta^2\right) &\beta\ll 1\\ \\ 1,496\beta^{-5/2}\left(1+5,107\beta^{-3/2}+14,43\beta^{-3}\right)&\beta\gg 1 \end{array}\right. $$ I know, how to get the first line: use Taylor series for $\sin(\beta x)$. But what should I do to prove the second line?

I tried to write the integral in the other form:

$$ H(\beta)=\frac{2}{\pi\beta}\int_{0}^{\infty}\exp\left(-\left(\frac{y}{\beta}\right)^{3/2}\right)\sin(y)ydy $$ and use Taylor series for $e^{-\left(y/\beta\right)^{3/2}}$, but faced with the divergent integral, which appears due to the first term of series:

$$\int_{0}^{\infty}\sin(y)ydy$$

Answer 1

Not an answer but too long for comments.

The first one is quite simple since $$H(\beta)=\frac{2}{\pi}\beta\int_{0}^{\infty}\exp\left(-x^{3/2}\right)\sin(\beta x)\,x\,dx$$ leads to $$\frac{729 \,\pi }{4 \,\beta ^2}H(\beta)=243 \, _3F_4\left(\frac{3}{4},1,\frac{5}{4};\frac{1}{3},\frac{2}{3},\frac{5}{6},\frac{7 }{6};-\frac{4 \beta ^6}{729}\right)+$$ $$28 \beta ^2 \Gamma \left(-\frac{2}{3}\right) \, _2F_3\left(\frac{13}{12},\frac{19}{12};\frac{2}{3},\frac{7}{6},\frac{3}{2};-\frac{4 \beta ^6}{729}\right)+$$ $$22 \beta ^4 \Gamma \left(\frac{2}{3}\right) \, _2F_3\left(\frac{17}{12},\frac{23}{12};\frac{4}{3},\frac{3}{2},\frac{11}{6};-\frac{4 \beta ^6}{729}\right)$$ where appear nasty hypergeomtric functions.

Developed as Taylor series built at $\beta=0$, this effectively gives $$H(\beta)=\frac{4 \beta ^2}{3 \pi }\left(1+\frac{28}{243} \Gamma \left(-\frac{2}{3}\right)\beta ^2+\frac{22}{243} \Gamma \left(\frac{2}{3}\right)\beta ^4+O\left(\beta ^6\right) \right)$$ and we could get as many terms as required.

For large values of $\beta$, I really do not see how the approximation could be made.

Are you sure that the second one is not the result of some curve fit ? In fact, I wonder what are these coefficients.

Is there a way to have a loook to the book ? If you want, send me the relevant pages as pdf files (my e-mail address is in my profile). I would have a look.

Edit

Thanks to @Maxim's comment, the problem is clarified. Using $x=\frac y \beta$ and expanding the exponential term, we end with $$H( \beta)=\frac 2 {\pi \beta}\int_0^\infty \sum_{n=0}^\infty (-1)^n\frac{ y^{\frac{3 n}{2}+1} \sin (y)}{\beta^{\frac{3n}2} n!}\,dy$$ and $$\int_0^\infty y^{\frac{3 n}{2}+1} \sin (y)\,dy=-\sin \left(\frac{3 \pi n}{4}\right) \Gamma \left(\frac{3 n+4}{2}\right)\qquad \text{if}\qquad \color{red}{ -2<\Re(n)<-\frac{2}{3}}$$

Ignoring the condition, this would lead to $$H(\beta)=\frac{15}{4 \sqrt{2 \pi } }\beta ^{-5/2}\left(1+\frac{32}{5} \sqrt{\frac{2}{\pi }}\beta ^{-3/2} +\frac{231}{16 }\beta ^{-3}-\frac{153153}{512 }\beta ^{-6}+O(\beta)^ {-15/2 } \right)$$

Answer 2

Well, we are looking (in a more general sense) at the following integral:

$$\mathcal{H}_\text{n}\left(\beta\right):=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty x\cdot\sin\left(\beta\cdot x\right)\cdot\exp\left(-x^\text{n}\right)\space\text{d}x\tag1$$

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:

$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\mathcal{L}_x\left[x\cdot\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-x^\text{n}\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$

Using the 'frequency-domain derivative' property of the Laplace transform we can write:

$$\mathcal{L}_x\left[x\cdot\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}=-\frac{\partial}{\partial\sigma}\left\{\mathcal{L}_x\left[\sin\left(\beta\cdot x\right)\right]_{\left(\sigma\right)}\right\}=\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\tag3$$

So:

$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\cdot\mathcal{L}_x^{-1}\left[\exp\left(-x^\text{n}\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag4$$

Finding the other functions, we can write:

$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{2}{\pi}\cdot\beta\cdot\int_0^\infty\frac{2\cdot\beta\cdot\sigma}{\left(\sigma^2+\beta^2\right)^2}\cdot\left\{\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{\sigma^{-1-\text{k}\text{n}}}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\right\}\space\text{d}\sigma=$$ $$\frac{4\cdot\beta^2}{\pi}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\left\{\int_0^\infty\frac{\sigma^{-\text{k}\text{n}}}{\left(\sigma^2+\beta^2\right)^2}\space\text{d}\sigma\right\}\tag5$$

Using Mathematica I found that:

$$\int_0^\infty\frac{\sigma^{-\text{k}\text{n}}}{\left(\sigma^2+\beta^2\right)^2}\space\text{d}\sigma=\frac{\pi}{4}\cdot\frac{1+\text{kn}}{\beta^{3+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\tag6$$

So, in the end we get:

$$\mathcal{H}_\text{n}\left(\beta\right)=\frac{4\cdot\beta^2}{\pi}\cdot\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\left\{\frac{\pi}{4}\cdot\frac{1+\text{kn}}{\beta^{3+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\right\}=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\text{n}\right)}\cdot\frac{1+\text{kn}}{\beta^{1+\text{kn}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{kn}\right)\tag7$$

In your example we have $\text{n}=\frac{3}{2}$, so:

$$\mathcal{H}_\frac{3}{2}\left(\beta\right)=\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\frac{3}{2}\right)}\cdot\frac{1+\text{k}\cdot\frac{3}{2}}{\beta^{1+\frac{3}{2}\text{k}}}\cdot\sec\left(\frac{\pi}{2}\cdot\text{k}\cdot\frac{3}{2}\right)=$$ $$\sum_{\text{k}=0}^\infty\frac{\left(-1\right)^\text{k}}{\text{k}!}\cdot\frac{1}{\Gamma\left(-\text{k}\cdot\frac{3}{2}\right)}\cdot\frac{1+\text{k}\cdot\frac{3}{2}}{\beta^{1+\frac{3}{2}\text{k}}}\cdot\sec\left(\frac{3\pi}{4}\cdot\text{k}\right)\tag8$$

Answer 3

We can also use the Laplace method to derive the asymptotic expansion of the integral. To isolate the large parameter in the exponent, we can rewrite the integral by changing $x=\beta^2u^2$: \begin{align} H(\beta)&=\frac{2}{\pi}\beta\int_{0}^{\infty}\exp\left(-x^{3/2}\right)\sin(\beta x)x\,dx\\ &=\frac{2}{\pi}\beta\Im\int_{0}^{\infty}\exp\left(-x^{3/2}+i\beta x\right)x\,dx\\ &=\frac{4}{\pi}\beta^5\Im\int_{0}^{\infty}\exp\left(-\beta^3\left( u^{3}-iu^2\right)\right)u^3\,du \end{align} The function $p(u)=u^{3/2}-iu^2$ is such that $\Re\left[ p(u)\right] >0$ for $u>0$ and $\Re\left[ p(0)\right] =0$. Then we can use the Laplace method to evaluate the large $\beta$ behaviour of the integral which is controlled by the the vicinity of $u=0$. With the notation of the above link, \begin{align} z&=\beta^3\\ p(u)&=-iu^2+u^3;\quad \mu=2,\quad p(0)=0\\ q(u)&=u^3;\quad \lambda=4 \end{align} The expansion reads \begin{align} H(\beta)\sim\frac{4}{\pi}\beta^5\Im\sum_{s=0}^\infty\Gamma\left( \frac{s+4}{2} \right)b_s\beta^{-\frac{3\left( s+4 \right)}{2}} \end{align} where \begin{align} b_s&=\frac{1}{2}\operatorname{Res}\left[ \frac{u^3}{\left( -iu^2+u^3 \right)^{\frac{s+4}{2}}}\right]_{u=0}\\ &=\frac{1}{2}\operatorname{Res}\left[ \frac{u^{-1-s}}{\left( u-i \right)^{\frac{s+4}{2}}}\right]_{u=0}\\ &=\frac{1}{2s!}\frac{d^s}{du^s}\left[\frac{1}{\left(u-i \right)^{\frac{s+4}{2}}}\right]_{u=0}\\ &=\frac{(-1)^s}{2s!}\frac{\Gamma\left( \frac{3s+4}{2} \right)}{\Gamma\left( \frac{s+4}{2}\right)}\left( -i \right)^{{-\frac{3s+4}{2}}} \end{align} Then, \begin{align} H(\beta)&\sim\frac{2}{\pi}\Im\sum_{s=0}^\infty\frac{(-1)^s}{s!}\Gamma\left( \frac{3s+4}{2} \right)\beta^{-\frac{\left( 3s+2 \right)}{2}}\left( -i \right)^{-\frac{3s+4}{2}}\\ &\sim\frac{2}{\pi}\sum_{s=1}^\infty\frac{(-1)^{s+1}}{s!}\Gamma\left( \frac{3s+4}{2} \right)\beta^{-\frac{\left( 3s+2 \right)}{2}}\sin\frac{3s\pi}{4} \end{align} which is the result quoted in the link given by @olexot in the comment.

Answer 4

We can apply the steepest descent method to $\exp(-x^{3/2} + i \beta x)$. It can be proved that it's sufficient to take an integration contour that goes in the direction $i$ from the endpoint $x = 0$ and estimate the integral over a small interval near the endpoint (outside of the interval the contour can be taken to approach the real axis). Then the part of the integrand that does not depend on $\beta$ can be expanded into series and the integration range can be extended to infinity, giving $$\int_0^\infty x e^{-x^{3/2} + i \beta x} dx \sim -\int_0^\epsilon \xi e^{-(i \xi)^{3/2} - \beta \xi} d\xi \sim -\sum_{k \geq 0} \int_0^\infty \frac {(-(i \xi)^{3/2})^k} {k!} \xi e^{-\beta \xi} d\xi,$$ where $\sim$ means that this is a complete (infinite) asymptotic series for both integrals. Taking the imaginary part and multiplying by $2 \beta/\pi$ gives $$\frac 2 \pi \sum_{k \geq 0} \frac {\sin \frac {\pi k} 4} {k!} \Gamma {\left( \frac {3 k} 2 + 2 \right)} \beta^{-3 k/2 - 1}.$$