Show integral is equal to Bessel function

  • Thread starter jackmell
  • Start date
  • #1
1,800
53
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:

Answers and Replies

  • #2
23
0
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
1,800
53
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.
 

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