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

[oh20elyf]

I am struggling to find the antiderivative of the following function:f(x)=\frac{J_{0}(ax)J_{1}(bx) }{x+x^{4} } <br /> \\<br /> J_{0},{~}J_{1} : Bessel{~}function{~}of{~}the{~}first{~}kind\\<br /> a, b: constants<br /> \\<br /> F(x)=\int_{}^{} \! f(x) \, dx =?Who can help?

 

[alyafey22]

I am struggling to find the antiderivative of the following function:f(x)=\frac{J_{0}(ax)J_{1}(bx) }{x+x^{4} } <br /> \\<br /> J_{0},{~}J_{1} : Bessel{~}function{~}of{~}the{~}first{~}kind\\<br /> a, b: constants<br /> \\<br /> F(x)=\int_{}^{} \! f(x) \, dx =?Who can help?

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

 

[oh20elyf]

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:

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

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:

\frac{d^2d(x)}{dx^2}=?

 

[Ackbach]

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.