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Re: a definite integral

• To: mathgroup at smc.vnet.net
• Subject: [mg77115] Re: a definite integral
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Mon, 4 Jun 2007 03:47:37 -0400 (EDT)
• References: <f3u421$2nq$1@smc.vnet.net>
• Reply-to: "Dr. Wolfgang Hintze" <weh at snafu.de>

"dimitris" <dimmechan at yahoo.com> schrieb im Newsbeitrag 
news:f3u421$2nq$1 at smc.vnet.net...
> dint=Integrate[BesselJ[0, x]BesselJ[0, t - x], {x, 0, t}]
>
> Mathematica (5.2) returns unevaluated the integral.
>
> However the result is actually Sin[t].
>
> Check
>
> In[77]:=
> o = Table[Random[Real, {1, 10}], {10}]
> Out[77]=
> {5.99068,7.64822,8.52201,6.97126,5.02596,7.44021,9.8316,5.03409,8.95359,1.\
> 34601}
>
> In[78]:=
> (NIntegrate[BesselJ[0, u]*BesselJ[0, #1 - u], {u, 0, #1}] & ) /@ o
> Out[78]=
> {-0.288349,0.978906,0.785043,0.635047,-0.951238,0.915609,-0.395691,-0.948699,\
> 0.453948,0.974842}
>
> In[79]:=
> Sin[o]
> Out[79]=
> {-0.288349,0.978906,0.785043,0.635047,-0.951238,0.915609,-0.395691,-0.948699,\
> 0.453948,0.974842}
>
> Any ideas to show dint is equal to sin(t)?
> By hand this can be proved by the convolution theorem
> of the LaplaceTransform.
>
> Dimitris
>
>
In 5.2 you could use Series to develop one Bessel function into a power 
series and integrate term by term after which you can recognize Sin[t]:
In[3]:=
Integrate[BesselJ[0, t - x]*Series[BesselJ[0, x],
{x, 0, 11}], {x, 0, t}]
Out[3]=
SeriesData[t, 0, {1, 0, -1/6, 0, 1/120, 0, -1/5040, 0,
1/362880, 0, -1/39916800}, 1, 13, 1]

Regards,
Wolfgang