Re: Can this integration be done?

*To*: mathgroup at smc.vnet.net*Subject*: [mg50116] Re: [mg50074] Can this integration be done?*From*: DrBob <drbob at bigfoot.com>*Date*: Sun, 15 Aug 2004 03:15:13 -0400 (EDT)*References*: <200408140550.BAA15300@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Question: what's the analytic expression for Integrate[1/t,{t,1,x}]? If you say Log[x] you're half right, but what IS Log[x], and how do you compute it? There actually is no closed-form expression for Integrate[1/t,{t,1,x}]; we've simply NAMED it Log[x]. So... that's a very simple integral, isn't it? Why should YOUR integral have a closed-form solution? Here's an antiderivative (apparently): antiderivative = Integrate[x/((a + x)*(b + (c + x^2)^2)), x] (-2*(-1)^(1/4)*(Sqrt[b] - I*c)*Sqrt[Sqrt[b] + I*c]* (I*a^2 + Sqrt[b] + I*c)* ArcTan[((-1)^(3/4)*x)/ Sqrt[Sqrt[b] - I*c]] - Sqrt[Sqrt[b] - I*c]* (2*(-1)^(1/4)*(Sqrt[b] + I*c)*(a^2 + I*Sqrt[b] + c)* ArcTan[((-1)^(1/4)*x)/ Sqrt[Sqrt[b] + I*c]] + a*Sqrt[Sqrt[b] + I*c]* (2*(a^2 + c)*ArcTan[ Sqrt[b]/(c + x^2)] + Sqrt[b]* (4*Log[a + x] - Log[b + (c + x^2)^ 2]))))/(4*Sqrt[b]* Sqrt[Sqrt[b] - I*c]* Sqrt[Sqrt[b] + I*c]* (a^4 + b + 2*a^2*c + c^2)) If you compute this at x = 0 and Infinity, then subtract, does that give the right integral? It depends on whether there's a "branch" of the antiderivative that's continuous on the positive real line. But that depends, in a complicated way, on the values of a, b and c. Look at the Log and ArcTan terms, and consider all the possible branch cuts you might run into. Still, you're welcome to subtract these two results: antiderivative /. {x -> 0} -((a*(2*(a^2 + c)*ArcTan[ Sqrt[b]/c] + Sqrt[b]*(4*Log[a] - Log[b + c^2])))/ (4*Sqrt[b]*(a^4 + b + 2*a^2*c + c^2))) Limit[antiderivative, x -> Infinity] -((I*((-I)*a^2*Sqrt[b]* Sqrt[Sqrt[b] + I*c]*Pi* Sqrt[-(I/Sign[Sqrt[b] - I*c])]*Sqrt[ Sign[Sqrt[b] - I*c]] - b*Sqrt[Sqrt[b] + I*c]*Pi* Sqrt[-(I/Sign[Sqrt[b] - I*c])]*Sqrt[ Sign[Sqrt[b] - I*c]] - a^2*Sqrt[Sqrt[b] + I*c]* c*Pi*Sqrt[ -(I/Sign[Sqrt[b] - I*c])]*Sqrt[ Sign[Sqrt[b] - I*c]] - Sqrt[Sqrt[b] + I*c]*c^2* Pi*Sqrt[ -(I/Sign[Sqrt[b] - I*c])]*Sqrt[ Sign[Sqrt[b] - I*c]] - I*a^2*Sqrt[b]* Sqrt[Sqrt[b] - I*c]*Pi* Sqrt[I/Sign[Sqrt[b] + I*c]]*Sqrt[ Sign[Sqrt[b] + I*c]] + b*Sqrt[Sqrt[b] - I*c]*Pi* Sqrt[I/Sign[Sqrt[b] + I*c]]*Sqrt[ Sign[Sqrt[b] + I*c]] + a^2*Sqrt[Sqrt[b] - I*c]* c*Pi*Sqrt[ I/Sign[Sqrt[b] + I*c]]* Sqrt[Sign[Sqrt[b] + I*c]] + Sqrt[Sqrt[b] - I*c]*c^2* Pi*Sqrt[I/Sign[ Sqrt[b] + I*c]]* Sqrt[Sign[Sqrt[b] + I*c]]))/(4*Sqrt[b]* Sqrt[Sqrt[b] - I*c]* Sqrt[Sqrt[b] + I*c]* (a^4 + b + 2*a^2*c + c^2))) Neither is defined for all a, b, and c, even if branch cuts DON'T interfere. Bobby On Sat, 14 Aug 2004 01:50:24 -0400 (EDT), Daohua Song <ds2081 at columbia.edu> wrote: > Dear Group, > I am wondering whether the following integration has an analytic > expression. Alough i believe so. > Integrate[x/((a + x)*(b + (c + x^2)^2)),{0,Infinity}]. > When i set a=b=c=1,mathematica give me an answer. > > Any help is appreciated. > sincerely > Daohua > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Can this integration be done?***From:*Daohua Song <ds2081@columbia.edu>