Re: Numerical integration
- To: mathgroup at smc.vnet.net
- Subject: [mg73890] Re: Numerical integration
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Fri, 2 Mar 2007 06:47:13 -0500 (EST)
- References: <es3iul$obt$1@smc.vnet.net><es6dku$s5p$1@smc.vnet.net>
What version do you use? Hello Daniel and thanks for your response! I didn't notice that for a>1 and b>1 Integrate returns a symbolic result. See my relevant post here: http://groups.google.gr/group/comp.soft-sys.math.mathematica/browse_thread/= thread/00897ab3139d7d82/dbc16a40e8ca0a1a?hl=el#dbc16a40e8ca0a1a Anyway doesn't it look strange that? Integrate[Cos[a*x]*CosIntegral[b*x], {x, 0, Infinity}] Integrate::idiv : Integral of Cos[a\x]\CosIntegral[b\x] does not converge on \ {0, =E2=88=9E}. Integrate[Cos[a*x]*CosIntegral[b*x], {x, 0, Infinity}] Integrate[Cos[a*x]*CosIntegral[b*x], {x, 0, Infinity}, GenerateConditions -> False] Integrate[Cos[a*x]*CosIntegral[b*x], {x, 0, Infinity}, GenerateConditions -> False] BUT Integrate[Cos[a*x]*CosIntegral[b*x], {x, 0, Infinity}, Assumptions -> a > 1 && b > 1] -((Pi*(a - b + Abs[a - b]))/(4*a*(a - b))) Dimitris =CE=9F/=CE=97 dh =CE=AD=CE=B3=CF=81=CE=B1=CF=88=CE=B5: > Hi Dimitris, > > why do you want to calculate numerically an integral that can be done > > analytically? For a,b >1 we get: > > > > -(Pi (a - b + Abs[a - b])) > > -------------------------- > > 4 a (a - b) > > Daniel > > > > dimitris wrote: > > > In another post I talk about the integral > > > > > > Integrate[Cos[a x] CosIntegral[b x], {x, 0, Infinity}] > > > > > > I have problems to numerical integrate this function for say > > > {a,b}={3,2}. > > > > > > In[20]:= > > > Integrate[Cos[3*x]*CosIntegral[2*x], {x, 0, Infinity}] > > > N@% > > > > > > Out[20]= > > > -(Pi/6) > > > Out[21]= > > > -0.5235987755982988 > > > > > > No matter how I set Options I couldn't get satisfactory results by > > > NIntegrate. > > > > > > Any ideas will be greatly appreciate! > > > > > > Here is its plot > > > > > > In[59]:= > > > Plot[Cos[3*x]*CosIntegral[2*x], {x, 0, 10}, Ticks -> {Range[0, 10*Pi, > > > Pi/6], Automatic}] > > > > > > As we see the zeros if the function are situated at Pi/6 + n*(Pi/3), > > > n=0,1,2,3... > > > > > > In[61]:= > > > (Cos[3*#1]*CosIntegral[2*#1] & ) /@ Table[Pi/6 + n*(Pi/3), {n, 0, > > > 100}] > > > > > > Out[61]= > > > {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,= 0,0,0,\ > > > 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0= ,0,0,0,\ > > > 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0} > > > > > > I tried to take use of this fact doing something like > > > > > > In[67]:= > > > lst = Table[Pi/6 + n*(Pi/3), {n, 0, 1000}] /. {a_, b__, c_} -> {x, 0, > > > a, b, c}; > > > > > > In[70]:= > > > NIntegrate[Cos[3*x]*CosIntegral[2*x], Evaluate[Sequence[lst]], > > > WorkingPrecision -> 40] > > > NIntegrate::ncvb :.... > > > -0.52359885758572151495786704 > > > > > > Very good result but I look for any other methods/settings! > > > > > > Dimitris > > > > > >