       Re: Re: Bug? Analytical integration of cosines gets the

• To: mathgroup at smc.vnet.net
• Subject: [mg107365] Re: [mg107346] Re: Bug? Analytical integration of cosines gets the
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Thu, 11 Feb 2010 05:16:05 -0500 (EST)
• References: <hkeb9k\$b5\$1@smc.vnet.net> <201002100835.DAA21213@smc.vnet.net>

```That's a VERY simple integrand; for Mathematica to get it wrong is
worrisome.

Below is an analytical confirmation of the NIntegrate result.

expr = Cos[ph]*1/Pi*Cos[4*ph]*Cos[2*ph];
Integrate[expr, {ph, Pi, 3/2*Pi}]

19/(105 \[Pi])

expr2 = Cos[ph]*1/Pi*Cos[4*ph]*Cos[2*ph] // TrigReduce
Integrate[expr2, {ph, Pi, 3/2*Pi}]

(Cos[ph] + Cos[3 ph] + Cos[5 ph] + Cos[7 ph])/(4 \[Pi])

-(19/(105 \[Pi]))

Bobby

On Wed, 10 Feb 2010 02:35:02 -0600, WetBlanket <wyvern864 at gmail.com> wrote:

> On Feb 4, 5:33 am, K <kgs... at googlemail.com> wrote:
>> Hello everyone,
>>
>> the analytical integration in Mathematica 7.01.0 on Linux x86 (64bit)
>>
>> faultyInt =
>>  Integrate[Cos[ph]*1/Pi*Cos[4*ph]*Cos[2*ph], {ph, Pi, 3/2*Pi}]
>>
>> gives as result:
>>
>> 19/(105 \[Pi])
>>
>> which is as a decimal number
>>
>> N[faultyInt]
>>
>> 0.0575989
>>
>> The numerical integration
>>
>> NIntegrate[Cos[ph]*1/Pi*Cos[4*ph]*Cos[2*ph],{ph,Pi,3/2*Pi}]
>>
>> gives
>>
>> -0.0575989
>>
>> which I believe is correct by judging from the plot
>>
>> Plot[Cos[ph]*1/Pi*Cos[4*ph]*Cos[2*ph], {ph, Pi, 3/2*Pi},
>>  PlotRange -> {-1/Pi, 1/Pi}]
>>
>> and because the quadgk function in another system gives the same
>> negative result.  Could anyone try this at home (or work, rather)
>> and confirm or disprove it?
>> Thanks,
>> K.
> If One substitutes the sequence, {1.9, 1.99, 1.999, ... 2.0}
> Mathematica gets the correct answer for as long as I tried except, of
> course for 2.0.
>

--
DrMajorBob at yahoo.com

```

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