Integral of the Von karman equation

Question

What is the result of this integral, and how can I proceed:

$$ \int_{-\infty}^{\infty}{c_{1} \over\left(1 + c_{2}\,x^{2}\right)^{5/6}}\, \cos\left(x\tau\right)\,{\rm d}x\,,\qquad c_{1}, c_{2}\mbox{: positive constants.} $$

I see in one book that the result contains the ${\tt 2_{\rm nd}\ \mbox{type Bessel function}}$.

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

Related problems (I). Follow the steps

1) the integrand is even so you can integrate on the interval $(0,\infty)$

2) expand the function $\cos(x)$ in terms of its Taylor series

3) use the beta function to evaluate the integral and then resum.

Added: Here is a closed form for a more general integral

$$ c_1\int_{-\infty}^{\infty}{\frac{\cos\left(x\tau\right)}{\left(1 + c_{2}\,x^{2}\right)^{\alpha}}}\, \,{\rm d}x = \frac{ c_{{1}}{c_{{2}}}^{-1/4-1/2\,\alpha}{\pi}^{3/2} \,{\tau}^{\alpha-1/2}{2}^{1/2-\alpha} }{\Gamma \left(\alpha\right) \cos \left( \pi \,\alpha \right) } {{\rm I}_{1/2-\alpha}\left({\frac { {\tau}}{\sqrt {c_{{2}}}}}\right)} $$

in terms of the modified Bessel function of the first kind. Pay attention for what values of $\alpha$ the above formula make sense.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 12 '14 at 19:05

Mhenni Benghorbal

47,431

if we have only x and not x^2. How can I proceed? – Ahmed Abrous Aug 14 '14 at 16:23
@AhmedAbrous: I think I gave you the steps how to solve this problem and I gave you a formula for a general case! – Mhenni Benghorbal Aug 14 '14 at 16:53
@ Mhenni Benghorbal Yes It is well answered, but I can't accept all of the two answers. I don't know why. The second question is that I suppose to have only x at the place of $x^2$. Thank you alot. – Ahmed Abrous Aug 14 '14 at 16:55
@AhmedAbrous: Still you can get a solution if you replace $x^2$ by $x$. – Mhenni Benghorbal Aug 14 '14 at 17:02
@AhmedAbrous: Try to work out the problem. I'll be with you later. – Mhenni Benghorbal Aug 14 '14 at 17:04
1

Good, I will try to do the same thing with Taylor series, and I tell you what it gives . – Ahmed Abrous Aug 14 '14 at 17:07
1

@AhmedAbrous: See what I added. – Mhenni Benghorbal Aug 14 '14 at 19:45
Let us continue this discussion in chat. – Ahmed Abrous Aug 14 '14 at 20:32
@ Mhenni Benghorbal The result that you gave me contains the Lommel function which tends to zero, but When we have treated the same problem with Mellin Transform, we have not find the same thing. – Ahmed Abrous Aug 15 '14 at 00:10
@AhmedAbrous: What do you mean by "tends to $0$"? – Mhenni Benghorbal Aug 15 '14 at 00:23
When $\tau$ tends to infinity, then the Integral tends to 0, but with the Mellin Transform It is not the same result. Can I send you the development by mail? – Ahmed Abrous Aug 15 '14 at 00:26
@AhmedAbrous: Are you interested in asymptotic behavior of the integral for large $\tau$? – Mhenni Benghorbal Aug 15 '14 at 00:28
it's done. Thank you. – Ahmed Abrous Aug 15 '14 at 00:56
@AhmedAbrous: I'll add some material. – Mhenni Benghorbal Aug 15 '14 at 00:59
Why In my case of solution that I sent you, I don't have the same thing? – Ahmed Abrous Aug 15 '14 at 01:44
@AhmedAbrous: I made a small correction. – Mhenni Benghorbal Aug 15 '14 at 03:11
@AhmedAbrous: It is $1/(1+c_2x)^{5/6}$ not $1/(1+c_2x)^{5/3}$. What you sent me is $1/(1+c_2x)^{5/3}$ – Mhenni Benghorbal Aug 15 '14 at 03:21
I will send the other integral soon. – Ahmed Abrous Aug 15 '14 at 13:46
@AhmedAbrous: It is better to post a new question for the other problem. – Mhenni Benghorbal Aug 15 '14 at 22:50
I think that the result is not good. I send you the detail. – Ahmed Abrous Aug 22 '14 at 20:53
1

@AhmedAbrous: the most important thing was given to you which is the techniques! The rest is your job. However I will double check my answer. – Mhenni Benghorbal Aug 22 '14 at 21:39

Felix Marin · Answer 2 · 2018-04-13T21:41:48.073

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\, \cos\pars{x\tau}\,\dd x\,,\qquad c_{1}, c_{2}\mbox{: positive constants.}}$

\begin{align}&\int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\, \cos\pars{x\tau}\,\dd x =2c_{1}\int_{0}^{\infty}{\cos\pars{x\verts{\tau}} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\, \,\dd x \\[3mm]&={2c_{1} \over c_{2}^{5/6}}\int_{0}^{\infty} {\cos\pars{x\color{#c00000}{\verts{\tau}}}\over \bracks{x^{2} + \pars{\color{#c00000}{c_{2}^{-1/2}}}^{2}}^{\color{#c00000}{1/3} + 1/2}}\,\dd x \end{align}

With the Bessel Function Identity ${\bf 9.6.25}$: $$ {\rm K}_{\nu}\pars{xz} ={\Gamma\pars{\nu + 1/2}\pars{2z}^{\nu} \over \root{\pi}x^{\nu}} \int_{0}^{\infty}{\cos\pars{xt} \over \pars{t^{2} + z^{2}}^{\nu + 1/2}}\,\dd t $$ we'll have \begin{align} &\int_{-\infty}^{\infty}{\cos\pars{x\tau} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\, \,\dd x ={\root{\pi}\verts{\tau}^{1/3} \over \Gamma\pars{1/3 + 1/2}\pars{2/\root{c_{2}}}^{1/3}}\, {\rm K}_{1/3}\pars{\verts{\tau}\,{1 \over \root{c_{2}}}} \end{align}

\begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\, \cos\pars{x\tau}\,\dd x} \\[3mm]&=\color{#66f}{\large{c_{1}\root{\pi}\over \Gamma\pars{5/6}}\,\pars{\verts{\tau}\root{c_{2}} \over 2}^{1/3} \,{\rm K}_{1/3}\pars{\verts{\tau} \over \root{c_{2}}}} \end{align}

A great answer (+1)! (I know $c_1$ is irrelevant here, but shouldn't it be in the final expression?) — Yuriy S, Apr 13 '18 at 21:21

Integral of the Von karman equation

2 Answers2

Linked