MHB How Do You Integrate a Product of Bessel Functions with Polynomial Denominators?

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The discussion focuses on finding the antiderivative of a function involving Bessel functions and polynomial denominators. The user is specifically trying to compute the integral of a function that relates to pavement deflections from traffic loads. They seek a method to derive the second derivative with respect to 'x' for acceleration calculations. A key suggestion is to differentiate the first Bessel function within the integral directly, as 'x' only appears there and not in the limits of integration. This approach simplifies the process of obtaining the desired result.
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I am struggling to find the antiderivative of the following function:[tex]f(x)=\frac{J_{0}(ax)J_{1}(bx) }{x+x^{4} }
\\
J_{0},{~}J_{1} : Bessel{~}function{~}of{~}the{~}first{~}kind\\
a, b: constants
\\
F(x)=\int_{}^{} \! f(x) \, dx =?[/tex]Who can help?
 
Last edited:
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oh20elyf said:
I am struggling to find the antiderivative of the following function:[tex]f(x)=\frac{J_{0}(ax)J_{1}(bx) }{x+x^{4} }
\\
J_{0},{~}J_{1} : Bessel{~}function{~}of{~}the{~}first{~}kind\\
a, b: constants
\\
F(x)=\int_{}^{} \! f(x) \, dx =?[/tex]Who can help?

What is the context of the question? Should the answer be in terms of the Bessel function?
 
ZaidAlyafey said:
What is the context of the question? Should the answer be in terms of the Bessel function?

Ok so the root of my problem is that i have a formula to calculate pavement deflections from traffic loads, which is the following:

[tex]d(x)=\int_0^\infty \! \frac{J_{0}(atx)J_{1}(bt)}{t+t^{4} } \, d t[/tex]

a,b: constants
x: distance from load centre

Now i want to find a formula to calculate the acceleration. I have to find the second derivative of the above formula with respect of 'x'.

To do so i thought i had to get rid of the integral at first, by finding the antiderivative with respect of 't' of the equation first to be able to differentiate with respect of 'x' twice. Is there a better way to find:

[tex]\frac{d^2d(x)}{dx^2}=?[/tex]
 
Last edited:
Well, the only place the $x$ shows up is in the first Bessel function inside the integral. It's not even present in the limits of the integral. So, just differentiate that first Bessel function twice with respect to $x$, and you're done.
 

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