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Bessel functions in terms of the confluent hypergeometric function

I am familiar with the representation of the Bessel functions as

$$J_m(x) = left(frac{x}{2}right)^m sum_{k = 0}^{infty} frac{(-1)^k}{k! (k + m)!} left(frac{x}{2}right)^{2k},$$

for some integer $m$.

Now, I have come across the representation of the Bessel functions in terms of the confluent hypergeometric function as

$$J_m(x) = frac{1}{Gamma(m + 1)} left(frac{x}{2}right)^m e^{-i x} Phileft(m + frac{1}{2}, 2m + 1; 2i xright).tag{*}$$

I would like to show that these two are equivalent.

We know that

$$Phi(alpha, gamma; x) = frac{Gamma(gamma)}{Gamma(alpha)} sum_{k = 0}^{infty} frac{Gamma(k + alpha)}{Gamma(k + gamma) Gamma(k + 1)} x^k.$$

Thus, $J_m(x)$ in terms of the confluent hypergeometric function is:

$$J_m(x) = frac{1}{m!} left(frac{x}{2}right)^m e^{-i x} frac{Gamma(2m + 1)}{Gamma(m + frac{1}{2})} sum_{k = 0}^{infty} frac{Gamma(k + m + frac{1}{2})}{Gamma(k + 2m + 1) Gamma(k + 1)} (2i x)^k.$$

By exploiting the following two relations

begin{align*}
Gamma(n + 1) &= n!,
\
Gammaleft(n + frac{1}{2}right) &= frac{(2n)!}{2^{2n} n!} sqrt{pi},
end{align*}

the above reduces to

$$J_m(x) = left(frac{x}{2}right)^m e^{-i x} sum_{k = 0}^{infty} frac{(2k + 2m)!}{2^{2k} (k + 2m)! (k + m)! k!} (2i x)^k.$$

From here, I don’t know how one could go further, for example, how to get rid of the exponential term and etc., and to reproduce the representation of the Bessel functions as given in the beginning of this post.

Mathematics Asked by user785957 on November 21, 2021

1 Answers

One Answer

One may deduce $(text{*})$ from the Poisson's integral representation for $Re m>-1/2$, $$J_m(x)=frac{left(frac{x}{2}right)^m}{Gammaleft(m+frac12right)sqrtpi}int_{-1}^{1}(1-t^2)^{m-1/2}e^{ixt},dt,tag{1}label{poisson}$$ for a proof, expand $e^{ixt}$ into a power series and integrate termwise, and the one for $$Phi(alpha;gamma;x)=frac{Gamma(gamma)}{Gamma(alpha)Gamma(gamma-alpha)}int_0^1 z^{alpha-1}(1-z)^{gamma-alpha-1}e^{xz},dz qquad(Regamma>Realpha>0),$$ obtained the same way. The RHS of $(text{*})$ is then $$frac{(x/2)^m e^{-ix}}{Gamma(m+1)}frac{Gamma(2m+1)}{Gamma^2(m+1/2)}int_0^1big(z(1-z)big)^{m-1/2}e^{2ixz},dz.$$ Using the duplication formula for $Gamma(2m+1)$ and substituting $z=(1+t)/2$, we get the RHS of eqref{poisson}.

Answered by metamorphy on November 21, 2021

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