How to calculate the analytical result of this singular integral? [duplicate]

Question

I'm trying to solve this integral:

$I(a,b)=\displaystyle{\int_1^{\infty }\dfrac{e^{-ax}}{\sqrt{x^2-1} (1+x\sqrt{1-b^2}) (x-1/\sqrt{1-b^2})} \, dx} \,\, \,\,(a \;\textrm{real>0} \, , 0<b<1)$

by varying b from 0 to 1.

Mathematica didn't calculate this integral. Maybe it is too complicated to be done.

In Mathematica input form:

Integrate[Exp[-a x]/(Sqrt[x^2 - 1](1+x Sqrt[1-b^2])(x-(1/Sqrt[1-b^2]))), {x,1,Infinity}]

If I simply enter that into Mathematica, it instantly returns the same expression. How do I go about this? Is there any tricks that can be applied? Is there a way to get the a symbolic result?

Thank's.

Answer 1

I can't find a way to integrate it symbolically, but here is a way to approximate it:

data = Flatten[Table[{a, b, Im@NIntegrate[ Exp[-a x]/(Sqrt[x^2 - 1] 
                            (1 + x Sqrt[1 - b^2]) (x - (1/Sqrt[1 - b^2]))),
                                        {x, 1, Infinity}]},
                     {a, 1, 20, .5}, {b, 2, 20, .5}], 1];
model = k2 Exp[-k3 a] Exp[-k4 b];
fit = FindFit[data, model, {k2, k3, k4}, {a, b}];
modelf = Function[{a, b}, Evaluate[model /. fit]];
Show[Plot3D[modelf[a, b], {a, 1, 20}, {b, 2, 20}, PlotRange -> All], 
     ListPointPlot3D[data, PlotStyle -> Red]]

Answer 2

I would think (having become cautious) that the integral is divergent since there is no prescription of how to circumvent the pole in the integrand at x = x0 = 1/Sqrt[1-b^2] which is > 1 for 0 < b < 1.

As it is the integrand becomes 1/(x-x0) times a finite factor so that the integral is logarithmically divergent.

Regards, Wolfgang