Prev: Fitch's Symbolic Logic format for Euclid's IP #152; 2nd ed; Euclid's..
Next: This Week's Finds in Mathematical Physics (Week 280)
From: Cary on 28 Sep 2009 01:33
On Sun, 27 Sep 2009 16:35:31 -0700 (PDT), Kenneth Bull
<kenneth.bull(a)gmail.com> wrote:

>I am trying to do definite integral of 1/sqrt(sinx) from 0 to pi/2,
>do I have to use an improper integral since sinx not defined at 0 ?

Given those limits, simplest perhaps is to recognize the integral as
the Wallis sine formula, given here (taking n = -1/2):
<http://mathworld.wolfram.com/WallisCosineFormula.html>

The same result also follows using the Beta function. See (14) at
<http://mathworld.wolfram.com/BetaFunction.html>

The value of the integral can be expressed as:
1/2 Beta(1/4, 1/2) = 1/2 sqrt(pi) Gamma(1/4) / Gamma(3/4)

As already given by David, this is 2.622057554...


Also, as Robert and David have said, the value can be expressed as an
elliptic integral, in this case a complete elliptic integral of the
first kind. Following the notation at
<http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html>
the value of the integral can be expressed as sqrt(2) K(k), where the
elliptic modulus k = 1/sqrt(2).


With the substitution u = pi/2 - x, the integral becomes
Int[0..pi/2] 1/sqrt[ cos(u) ] du

In that form, an evaluation of the integral is given by (11) to (26)
at <http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html>.

First  |  Prev  | 
Pages: 1 2
Prev: Fitch's Symbolic Logic format for Euclid's IP #152; 2nd ed; Euclid's..
Next: This Week's Finds in Mathematical Physics (Week 280)