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Re: Solving integrals
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68672] Re: Solving integrals
*From*: Peter Pein <petsie at dordos.net>
*Date*: Wed, 16 Aug 2006 03:35:58 -0400 (EDT)
*References*: <ebhi41$11p$1@smc.vnet.net> <ebpm4o$p9i$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Jens Benecke schrieb:
> Jens Benecke wrote:
>
>> Hello everybody,
>>
>> I am trying to solve this integral:
>>
>> f :== int ( cos(x) / sqrt[A=C2=B2=C2=B7sin=C2=B2(x/2) + (Bx)=C2=B2 =
>> ] dx, x==0..N*2*pi
>
> Hello,
>
> sorry, this post apparently got mangled ("²" aka "^2" wasn't recognized).
> Here is the integral again:
>
> f := int ( cos(x) / sqrt[ A*A*sin(x/2)*sin(x/2) + (Bx)^2] dx, x=0..N*2*pi
>
>
>> where N==(int)0..20, and A and B are small fixed values (around
>> 0.0001..0.05).
...
>> Thank you! :)
>
Hi Jens,
sorry, your integral does not converge on the interval [0, something],
because your integrand is c_(-1)*x^(-1) + c_(1)*x + ... near x=0:
expr = Cos[x]/Sqrt[(a*Sin[x/2])^2 + (b*x)^2];
Normal[Simplify[Series[expr, {x, 0, 1}], a > 0 && b > 0]]//InputForm
Out[2]//InputForm=
2/(Sqrt[a^2 + 4*b^2]*x) +
((-11*a^2 - 48*b^2)*x)/(12*(a^2 + 4*b^2)^(3/2))
Peter
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