I am stuck on one question with integral. Help me please to show that with $n=1$ the following is true $$ \int_{0}^{\infty}\left(\frac{2^n}{t^n}\left(\frac{t^n}{2^nn!}-\frac{1}{2^{n+2}}\frac{t^{n+2}}{1!(n+1)!}+\frac{1}{2^{n+4}}\frac{t^{n+4}}{2!(n+2)!}-\ldots\right)\right)^2tdt=2 $$

Thank you for your help.

  • 3
    $\begingroup$ I think that there is related to the Bessel functions $\endgroup$ – Pedro Tamaroff Jun 16 '12 at 19:32
  • $\begingroup$ Did you miss a $n!$ in the denominator of the first term? $\endgroup$ – user17762 Jun 16 '12 at 19:37
  • $\begingroup$ @Marvis I was going to ask the same. It seems so. $\endgroup$ – Pedro Tamaroff Jun 16 '12 at 19:38
  • $\begingroup$ Thank you. It should be $n!$ in the denominator of the first term. $\endgroup$ – Aleks.K Jun 16 '12 at 21:22

Let's consider Bessel functions of the first kind as suggested by Peter Tamaroff : $$J_n(t):=\left(\frac t2 \right)^n \sum_{k=0}^\infty \frac {\left(-\frac {t^2}4 \right)^k}{k!(n+k)!}=\frac{t^n}{2^nn!}-\frac{1}{2^{n+2}}\frac{t^{n+2}}{1!(n+1)!}+\frac{1}{2^{n+4}}\frac{t^{n+4}}{2!(n+2)!}-\ldots$$

then you want a derivation of : $$ \int_{0}^{\infty}\left(\frac{2^n}{t^n}J_n(t)\right)^2t\,dt=2 $$

A proof of the more general formula (supposing $\ \Re(\mu+\nu+1)>\Re(\lambda)>0$) : $$\int_0^\infty \frac {J_\mu(at)J_\nu(at)}{t^\lambda}\,dt=\frac{(\frac a2)^{\lambda-1}\Gamma(\lambda)\Gamma\left(\frac {\mu+\nu-\lambda+1}2\right)}{2\Gamma\left(\frac{\lambda+\nu-\mu+1}2\right)\Gamma\left(\frac{\lambda+\nu+\mu+1}2\right)\Gamma\left(\frac{\lambda-\nu+\mu+1}2\right)}$$ is in Watson's book 'Bessel functions' page 403.

With $a=1, \nu=\mu=n, \lambda=2n-1\ $ this becomes : $$\int_0^\infty \frac {J_n(t)^2}{t^{2n-1}}\,dt=\frac{(\frac 12)^{2n-2}\Gamma(2n-1)\Gamma\left(1\right)}{2\Gamma\left(n\right)\Gamma\left(2n\right)\Gamma\left(n\right)}=\frac 1{2^{2n-1}(2n-1)(n-1)!^2}$$

so that (for $n$ positive integer) : $$ \int_{0}^{\infty}\left(\frac{2^n}{t^n}J_n(t)\right)^2t\,dt=\frac 2{(2n-1)(n-1)!^2} $$


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