Try to solve this integral using Mathematica symbolic computation program.

Hello Mohammed: I don't have access to Mathematica. Instead I will prefer a Fortran program. 

Hello Amaechi J. Anyaegbunam, 

Thank You that You are thought of me. I send You any  results by http://integrals.wolfram.com/ and http://www.wolframalpha.com/ as a 2 files attachment.zip. I'm not sure I understood you correctly. Please, answer me.Your sincerely: Anna Tomova

Dear Anna :  Thanks for your response. However, the integration is to be carried out from 0 to infinity. Thus, the answer will be independent of x (my s). 

Hello Amaechi J. Anyaegbunam,

I send You 2 another files.zip with 2 surfaces. When alpha is in [0,2] and x in [0, infinity) this is a function of 2 variables. It is clear, that the problem is difficult. . Please, answer me.Your sincerely: Anna Tomova.

Dear Anna: Please examine the attached pdf file to know the type of answer I expect. It concerns Bycroft's integration of an infinite integral encountered in horizontal harmonic vibration of a rigid disc on a halfspace. 

Dear prof. Amaechi J. Anyaegbunam,

I send You as a file attachment.zip some results by ·http://www.wolframalpha.com/. It could be possible that they will add You in the parts with Taylor's series about the integrated function. Your sincerely, Anna Tomova.

Hello Anna:  I am unable to access your zip files. Could you provide your contribution in the form of separate doc, pdf, or jpg files. Thank you!

Dear prof. Amaechi J. Anyaegbunam,

I send You as 4 files attachment  some results by ·http://www.wolframalpha.com/. It could be possible that they will add You in the parts with Taylor's series about the integrated function. Your sincerely, Anna Tomova.

Hello Anna:

  Thanks for the doc files you provided. I have examined them and does not see how they can help me answer the question I asked. The integral in my question has f(s) in the denominator. Your files are merely examining a problem Bycroft solved in 1956. Thanks very much again for your efforts.

Dear prof. Amaechi J. Anyaegbunam,

Now I send You 1 file attachment.doc with the series for the first multiplier:  s*sqrt(s^2-b^2)*Bessel[0,a*s] and for the second multiplier: 1/((s^2-0.5)^2-s^2*((s^2-1)*(s^2-b^2))^(1/2)) around 0  and around infinity. You can multiply, very carefully, if they are convergent .  Please, remember about the roots of the denominator. They are shown in the second file attachment.doc . After the multiplication, I You can integrate and determine F and G the real and imaginary parts of the improper definite integral .Your sincerely, Anna Tomova.

Dear Anna

If I knew how to obtain the value of the improper intregal via contour integration I would not have asked the question. What I want is the expressions for F and G in whatever form. Thank you.  

Dear prof. Amaechi J. Anyaegbunam,

Only now understood you - you want to represent improper integral as a complex number in algebraic form. I offer you a presentation in the first attachment. As a second attachment I send You such a presentation for BYCROFT'S SOLUTION FORHORIZONTAL VIBRATION, but unfortunately only got just the imaginary part, why? I look forward to your response. Sincerely Anna Tomova.

Dear Anna

   I thank you for your latest answer that had 'hit the nail on the head.' Your solution for F for Bycroft's problem is in order as you might have suspected but it is in the form of infinite integrals which are each divergent as they stand. However, they can be manipulated into convergent integrals as I show in the attached docx file a solution I deduced a week ago. But the problem still remains that they differ from Bycroft's F that are expressed in in terms of finite integrals. You have now exposed the core of my question. By which technique do you obtain Bycroft's  from your own form? 

     Secondly, you have also provided an answer to my main question. There are minor problems with your answer. Your G has not accounted for the Rayleigh pole at s = sR that is the solution to f(s) = 0. sR varies from 1.045 to 1.144. Also the denominator of the integral from 0 to η is incorrect.

   Your expression for F is apparently correct . But it is a herculean task to evaluate the infinite integral from 1 to ∞ that has a pole at s = sR. Which is why it would be convenient if such a pathological integral could be converted to a finite integral. Infinite integrals involving Bessel functions are known to be extremely difficult to evaluate numerically. It can be shown that the integrand tends to Jo(as) as s tends to infinity which compounds the quadrature problem

Dear prof. Amaechi J. Anyaegbunam,,

Thank You for Your response. Looked attachment and not find a difference between my formula for the real part of the improper integral in 1956 (Bycroft's F). How he became the improper integral a form of one integral with final frontiers? Why do you think  my denominator for the imaginary part of your first integral is not correct? I changed 2 places characters in the denominator of  integrate function (which recording as the square root of two multiplers is not difference) and you forward it as a file attachment  I look forward to your response. Sincerely Anna Tomova.

Dear Anna,

   Thanks for your contribution. I would appreciate if you address me simply as Amaechi Anyaegbunam. I am not yet a full Professor. My apologies if I have offended you with this remark.

    (1) Bycroft's solution  does not contain a residue term because f(s) is absent in the denominator of any of his integrands.  For the poles at s = 1 and  η the residues are zero.

 (2) Consider the integral of s*sqrt(s2-η2)Jo(as)/f(s) from 0 to η. f(s)= (s2-0.5)2- s2√[(s2-η2)(s2-1)]. f(s) should then be i√(η2-s2)* i√(1-s2)  = -√[(s2-η2)(s2-1)], where i = √(-1).  Thus  for s = 0 to  η, f(s) = g(s) = (s2- 0.5)2 + s2√[(s2-η2)(s2-1)]. QED.Thus in the expression for G, f(s) should be replaced by g(s) in the integral from 0 to η.

    Find attached a reliable plot of λF (= f1) and λG (= f2) versus 'a'  for ν = 1/3 where λ = a/4/(1-ν) with ν = Poisson's ratio. η = sqrt[(1-2ν)/(2-2ν)].Could you reproduce these plots with your uncorrected formulas?

                                   Dear Amaechi Anyaegbunam,

Probably there are more expressions for interpret Your formulas. For example let us consider the integral of s*sqrt(s2-η2)Jo(as)/f(s) from 0 to η. f(s)= (s2-0.5)2- s2√[(s2-η2)(s2-1)].  Then we have  in the quadratic root √(-1)(η2-s2)(-1)(1-s2)= √(-1)^2[(-s2+η2)(-s2+1)], Thus for s = 0 to η, it is true that f(s) = g(s) = (s2- 0.5)2 - s2√[(-s2+η2)(-s2+1)]. The imaginary unit  i is not √(-1, it is the couple (0,1). √(-1,in C has 2 values: (0,1) and (0,-1), because  (0,-1)^2 = (-1,0) too. I look forward to your response. Sincerely Anna Tomova.