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:


  • 1
    $\begingroup$ The book in fact suggests exactly that: expanding the exponential and integrating term by term: $$\frac {2 \beta} \pi \int_0^\infty \frac {(-1)^k} {k!} x^{3 k/2 + 1} \sin \beta x \,dx = \frac {2 (-1)^{k + 1} \sin \frac {3 \pi k} 4} {\pi k!} \Gamma {\left( \frac {3 k} 2 + 2 \right)} \beta^{-3k/2 - 1},$$ which gives the correct asymptotic series, even though the integral doesn't exist for $k \geq 0$. $\endgroup$
    – Maxim
    Feb 17, 2019 at 2:44
  • $\begingroup$ If this problem is important for your work or research, we could build much better approximations. For example, around $\beta=0$ $$H(\beta)=\frac{4 \beta ^2}{3 \pi }\frac{1+\left(\frac{28 \Gamma \left(-\frac{2}{3}\right)}{243}-\frac{11 \Gamma \left(\frac{2}{3}\right)}{14 \Gamma \left(-\frac{2}{3}\right)}\right)\beta^2 } {1- \frac{11 \Gamma \left(\frac{2}{3}\right)}{14 \Gamma \left(-\frac{2}{3}\right)}\beta^2}$$ For $\beta=1$, this would give $0.269037$ while the first one you posted gives $0.227899$ while the exact solution is $0.271221$. The same could be done for large values of $\beta$. $\endgroup$ Feb 17, 2019 at 8:32
  • $\begingroup$ @ClaudeLeibovici, thank you for the further development! Although $H(\beta)$ is tabulated, I'm interesting in integration methods, not in numbers. I'm trying to understand all answers and going to ask some questions in comments $\endgroup$
    – Olexot
    Feb 17, 2019 at 9:24
  • $\begingroup$ To tell the truth, I am fascinated by function approximations ! Where are they tabulated ? Cheers. $\endgroup$ Feb 17, 2019 at 9:26
  • $\begingroup$ For example, in this article with help of numerical integration (Table 1, $r_0/\lambda=0$) $\endgroup$
    – Olexot
    Feb 17, 2019 at 10:01

4 Answers 4


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.


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 $$H(\beta)=\frac{15}{4 \sqrt{2 \pi } }\beta ^{-5/2}\left(1+\frac{32}{5} \sqrt{\frac{2}{\pi }}\beta ^{-3/2} +\frac{231}{16 }\beta ^{-3}-\frac{153153}{512 }\beta ^{-6}+O(\beta)^ {-15/2 } \right)$$

  • $\begingroup$ Thank you so much for your generosity! $\endgroup$
    – Akira
    Feb 16, 2019 at 15:09
  • $\begingroup$ @LeAnhDung. You are welcome ! I am really curious about the asymptotics for large $\beta$. $\endgroup$ Feb 16, 2019 at 15:10
  • $\begingroup$ I found it on the page number 17 here. There is also general expression through gamma function there $\endgroup$
    – Olexot
    Feb 16, 2019 at 15:55
  • $\begingroup$ Faced with lack of my knowledge, when I tried to calculate $\int_{0}^{\infty}y^{\frac{3n}{2}+1}\sin(y)dy$. The indefitine integral can be written as $F(x)+C$, where: $$F(x)=-\frac{1}{2i}\left[(-i)^{3n/2}\Gamma\left(\frac{3n}{2}+2,ix\right)-i^{3n/2}\Gamma\left(\frac{3n}{2}+2,-ix\right)\right]$$ (There are incomplete gamma functions with complex lower limit). It can be shown that $F(0)=\sin\left(\frac{3\pi n}{4}\right)\Gamma \left(\frac{3n}{2}+2\right)$. But why is $\lim_{x\rightarrow\infty}\Gamma \left(\frac{3n}{2}+2,-ix\right)=0$ right? (I didn't find something about it) $\endgroup$
    – Olexot
    Feb 17, 2019 at 10:37

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:

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



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:


enter image description here

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:

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

  • $\begingroup$ Thank you! I haven't yet figured out all solution steps, but one detail in the answer make me worried. Series starts with the term $\beta^{-1}$ instead of $\beta^{-5/2}$. Might the answer have a mistake? $\endgroup$
    – Olexot
    Feb 17, 2019 at 9:38
  • $\begingroup$ @Olexot Well, first of all you're welcome. I'm glad that I could be of any help. For your question, I checked my answer but I do not see any mistakes yet. $\endgroup$ Feb 17, 2019 at 9:46
  • 1
    $\begingroup$ @Olexot Note that this is also a method that relies on steps that are only formally correct and therefore needs to be justified. The inverse transform of $\exp(-x^n)$ doesn't exist for $n > 1$, because of superexponential growth of $\exp(-x^n)$ on the vertical line. The integral of $\sigma^{-k n}/(\sigma^2 + \beta^2)^2$ doesn't exist for $k \geq 1 \land n \geq 1$ because of the singularity at zero. The final result is correct as a limit: $1/\Gamma(-3 k/2) \to 0$ for $k \to 0$, $\sec(3 \pi k/4)/\Gamma(-3 k/2) \to 12/\pi$ for $k \to 2$. $\endgroup$
    – Maxim
    Feb 17, 2019 at 12:25
  • $\begingroup$ @Maxim Exactly, thanks for your input. If you want you can edit that information into my answer. $\endgroup$ Feb 17, 2019 at 12:26

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.


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 $$\frac 2 \pi \sum_{k \geq 0} \frac {\sin \frac {\pi k} 4} {k!} \Gamma {\left( \frac {3 k} 2 + 2 \right)} \beta^{-3 k/2 - 1}.$$

  • $\begingroup$ Thank you! Unfortunately, I can't understand the second sentence. If I doesn't make a mistake, steepest descent method gives a saddle point $x_0=-\frac{4}{9}\beta^2$ (because $\left (-x^{3/2}+i\beta x \right )'=0$ at $x_0$), where a direction of steepest descent is $\theta=\frac{\pi}{4}$ (because $\left (-x^{3/2}+i\beta x \right )''=\frac{9i}{8\beta}$ at $x_0$, and there should be $\Re\left [\left (-x^{3/2}+i\beta x \right )''\bigg|_{x_0}(x-x_0)^2/2 \right ]<0,\quad\Im\left [\left (-x^{3/2}+i\beta x \right )''\bigg|_{x_0}(x-x_0)^2/2 \right ]=0$) $\endgroup$
    – Olexot
    Feb 19, 2019 at 17:37
  • $\begingroup$ The contour doesn't have both endpoints at infinity, so we have to consider the contributions from the endpoints as well as from the saddle points. In this case we do not have to consider the saddle points at all. If we take the rest of the contour as $\gamma = [i \epsilon, \infty + i \epsilon)$, the integral over $\gamma$ will contain a factor of $e^{-\beta \epsilon}$ ($\epsilon$ is a fixed number), making the integral smaller than any term in the asymptotic series. $\endgroup$
    – Maxim
    Feb 19, 2019 at 18:21

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