Re: 1/(1 + t^p)

Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> writes:

"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> writes:


"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote in message
news:XJeki.25824$C96.21888@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Anyone know a closed form solution for

int(1/(1 + t^p),t=0..1) and p > 0?


p > 1

Also, I don't mind some constraints on p such it being an integer, even,
etc..

For each positive integer p there is a closed form solution. Namely, let
r_j = exp(i pi (2j-1)/p), j=1..p-1, be the roots of the polynomial t^p + 1.

Then we have the partial fraction expansion

1/(1+t^p) = - sum_{j=1}^{p-1} r_j/(p (t - r_j))

so the integral is - sum_{j=1}^{p-1} r_j ln(1 - 1/r_j)/p.

I have figured out that it is equivilent to

1/p*lerchphi(-1,1,1/p)

You could use series:

1/(1+t^p) = sum_{j=0}^infty (-1)^j t^(jp) for |t| < 1

Integrate for t from 0 to 1 and you get sum_{j=0}^infty (-1)^j/(jp+1)
which Maple evaluates as hypergeom([1,1/p],[1+1/p],-1). This is
equivalent to the LerchPhi form.

but this doesn't help me much ;/

I doubt there's any general form that's much simpler.

I take that back: Maple says it's also equivalent to

(Psi(1/2 + 1/(2 p)) - Psi(1/(2 p)))/(2 p)
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada V6T 1Z2
.

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