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