Show integral is equal to Bessel function

In summary, the following is true: if n=k, the RHS is equal to the Bessel function. However, if n=k+1, the LHS is equal to the RHS.
  • #1
jackmell
1,807
54
Hi guys,

I'm pretty sure the following is true but I'm stuck proving it:
[tex]
\begin{align*}
\frac{1}{2\pi}\int_{-1}^1 \left(\frac{e^{\sqrt{1-y^2}}}{\sqrt{1-y^2}}+\frac{e^{-\sqrt{1-y^2}}}{\sqrt{1-y^2}}\right) e^{iyx} dy&=\frac{1}{2\pi i}\mathop\oint\limits_{|t|=1} \exp\left\{\sqrt{x^2-1}(t-1/t)/2\right\}t^{-1}dt\\
&=J_0(\sqrt{x^2-1})
\end{align*}
[/tex]

I tried the substitutions:

[tex]y=i/2(t-1/t)[/tex]
[tex]y=1/2(t-1/t)[/tex]
[tex]y=i/2\sqrt{x^2-1}(t-1/t)[/tex]

but not getting it. Also, I think only the first substitution converts the domain of integration into the required circle around the origin but that 's not too clear to me as well. Anyone can make a suggestion what to try next? Not sure if I should have posted this in the homework section.

Edit: Ok, I made a mistake in the Bessel integral notation. It's z/2 and not iz/2 so I changed it above and will re-do my calculations. May change things for me.

Thanks,
Jack
 
Last edited:
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  • #2
You can expand the exponential term on the rhs as a product of two series:

[itex]\exp\left(\frac{\sqrt{x^2-1}}{2}\left(t-\frac{1}{t}\right)\right)=\sum_{n=0}^{\infty}\frac{(\sqrt{x^2-1})^n}{2^nn!}t^n\sum_{k=0}^{\infty}\frac{(\sqrt{x^2-1})^k}{2^kk!}(-t)^{-k}[/itex]

and so it becomes
[itex]\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(\sqrt{x^2-1})^{n+k}}{2^{n+k}n!k!}(-1)^k\frac{1}{2\pi i}\oint\frac{t^{n-k}}{t}dt[/itex]

The contour integral is now 1 if [itex]n=k[/itex] and 0 otherwise, giving the series
[itex]\sum_{n=0}^{\infty}\frac{(\sqrt{x^2-1})^{2n}}{2^{2n}n!n!}(-1)^n[/itex]
which is the series definition for [itex]J_{0}\left(\sqrt{x^2-1}\right)[/itex].
 
  • #3
Ok, but that only shows how the RHS is equal to the Bessel function. I was wanting to figure out how to show the LHS is equal to the RHS.
 

1. What is the definition of a Bessel function?

A Bessel function is a special type of mathematical function that is used to solve differential equations and model physical phenomena such as heat transfer and sound waves. It was first studied by the mathematician Daniel Bernoulli in the 18th century and is named after the mathematician Friedrich Bessel.

2. How is a Bessel function related to an integral?

The Bessel function is related to an integral through the integral representation of the function. This means that the Bessel function can be expressed as an integral of another function, making it useful for solving certain types of integrals.

3. What is the significance of the Bessel function in physics?

The Bessel function is significant in physics because it is often used to solve differential equations that arise in physical problems. It is particularly useful in problems involving circular or cylindrical symmetry, such as heat transfer in a cylindrical object or the behavior of sound waves in a circular pipe.

4. How is the Bessel function used in engineering?

The Bessel function is used in engineering to solve problems involving wave propagation, heat transfer, and other physical phenomena. It is also used in signal processing and image reconstruction, as well as in the design of antennas and other electromagnetic devices.

5. Are there different types of Bessel functions?

Yes, there are multiple types of Bessel functions, including the first kind (J), second kind (Y), modified Bessel functions (I and K), and spherical Bessel functions (j and y). Each type has its own unique properties and applications, but they are all related to each other through various mathematical relationships.

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