       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>
• 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

```