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I would like to evaluate the following integral: $$\int_{1}^{\infty} x^{-5/3} \cos\left((x-1) \tau\right) dx$$ I get the Integral by Maple and it gives the Lommel function. After that, I will search an asymptotic as $\tau$ goes to $\infty$.

Ahmed Abrous
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1 Answers1

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Related problems (I). Here is how you can evaluate the integral. Making the change of variables $u=x-1$ gives

$$I = \int_{0}^{\infty}\frac{\cos(\tau u)}{(1+u)^{5/3}}du. $$

Using the Taylor series of $\cos(\tau u)$ we get

$$ I = \sum_{k=0}^{\infty}\frac{(-1)^k \tau^{2k} }{(2k)!} \int_{0}^{\infty}\frac{u^{2k}}{(1+u)^{5/3}}du. $$

Making the change of variables $1+u=\frac{1}{t}$ casts the above integral in terms of the $\beta$ function and then resum the series and the result will follow in terms of Lommel function.

$$I= \frac{3}{2}-{\frac {27}{8}}\,{x}^{2}+{\frac {27}{8}}\,{x}^{\frac{7}{6}}s_{\frac{11}{6},\frac{1}{2}}( x ). $$

Another form for the solution can be obtained in terms of the hypergeometric function

$$ I = \frac{3}{2} {}_1F_2\left( 1;\frac{1}{6},\frac{2}{3}; -\frac{x^4}{4} \right) . $$

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Mhenni Benghorbal
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  • It's not what Maple gives me. So where is the problem :) – Ahmed Abrous Aug 18 '14 at 16:28
  • @AhmedAbrous: Which answer are you talking about? – Mhenni Benghorbal Aug 18 '14 at 19:19
  • I talk about the Integral here. I have found the same thing by calculus as you have done but with Maple it is not the same thing. Can you try please. Thank you. – Ahmed Abrous Aug 19 '14 at 00:19
  • @AhmedAbrous: These solution are equivalent to the ones which are given by Maple. You can use them instead. You should evaluate the integral with the beta function and resum and you will get the first answer. – Mhenni Benghorbal Aug 19 '14 at 00:24
  • Yes but the plot of the two functions are not the same. I have tried to get the Integral directly by Maple. Can you type the Integral directly on Maple. I see also that with Maple the Integral goes to zero as tau goes to infinity. It's not the same thing as by calculus. – Ahmed Abrous Aug 19 '14 at 00:29
  • I see also that with the beta function B(x,y), x and y are positive values, I think that the problem is here. If you want I send you the development – Ahmed Abrous Aug 19 '14 at 00:51
  • I found the difference. In the development we have done, it is LommelS2 and not LommelS1. – Ahmed Abrous Aug 19 '14 at 01:21
  • @AhmedAbrous: in the above two answers replace $x$ by $\tau$. – Mhenni Benghorbal Aug 19 '14 at 06:11
  • Yes but the problem is not here. With Maple when I do the sum like mentionned I obtain LommelS1 function but When I calculate the Integral with Maple it gives LommelS2 instead. – Ahmed Abrous Aug 19 '14 at 14:17
  • @AhmedAbrous: I do not know what is wrong with Maple's answer! Use the results I gave you which I derived using my own techniques as you see. By the way, Have you tried Mathematica to evaluate the integral? – Mhenni Benghorbal Aug 19 '14 at 14:43
  • No but I will try to do it :). When you write s, what function you mean. because I find two Lommel functions here http://mathworld.wolfram.com/LommelFunction.html – Ahmed Abrous Aug 19 '14 at 17:39
  • @AhmedAbrous: I already put a link. It is LommelS1. – Mhenni Benghorbal Aug 19 '14 at 18:44