Improper integral

[Jose.]

Hello everyone, I would really thank you for helping me with this integral;

\(\displaystyle \int_{0}^{\pi/4}{1/(\sqrt{cos2x})}\,dx\)

I have to establish its convergence and express its value as the Euler's Gamma function,

I rewrote the integrand as \(\displaystyle \int_{0}^{\pi/4}{1/(\sqrt{1-2sin^2x})}\,dx\),
then I did:

\(\displaystyle t=2sin^2x, dt=4sinxcosxdx ; dx=dt/\sqrt{8t-4t^2}\)

and it turns out \(\displaystyle \int_{0}^{1}{(1-t)^{-1/2}(t-t^2/2)^{-1/2}}\,dx\)

I am not sure whether it is the right change or not and I do not know how to continue

I appreciate your attention.

 

[Deleted member 4993]

Hello everyone, I would really thank you for helping me with this integral

\(\displaystyle \int_{0}^{pi/4}{1/(\sqrt{cos2x})}\,dx\)

I have to establish its convergence and express its value as the Euler's Gamma function

I rewrote the integrand as \(\displaystyle \displaystyle \int_{0}^{\frac{\pi}{4}}\frac{dx}{\sqrt{1-2sin^2x}}\)
then I did:
\(\displaystyle t=2sin^2x, dt=4sinxcosxdx ; dx=dt/\sqrt{8t-4t^2}\)

and it turns out \(\displaystyle \displaystyle \int_{0}^{1}\frac{dt}{2\sqrt{(1-t)(2t-t^2)}}\ \) [-1/2 is the exponent (I could not write it right)]

I am not sure whether it is the right change or not and I do not know how to continue

I appreciate your attention

This is a very innocent looking nasty integration - involving elliptic function and gamma function.


integral_0^(pi/4) 1/sqrt(cos(2 x)) dx = (K(1/2))/sqrt(2)~~1.31103

Use wolframalpha.com to check.

 

[Jose.]

We have not seen the elliptic function in class so I think that is not what the exercise is asking

We have just seen examples that through changes of variables you transform the Beta function in a trigonometric integral and you can approximate it with the formula:


\(\displaystyle B(p,q)=(\Gamma(p))(\Gamma(q)/{(\Gamma(p+q))}\) where B(p,q) is the Euler's Beta function

Thank you very much for your answer

 

[DrPhil]

Hello everyone, I would really thank you for helping me with this integral;

\(\displaystyle \int_{0}^{\pi/4}{1/(\sqrt{cos2x})}\,dx\)

I have to establish its convergence and express its value as the Euler's Gamma function,

I rewrote the integrand as \(\displaystyle \displaystyle \int_{0}^{\pi/4}{1/(\sqrt{1-2sin^2x})}\,dx\),
then I did:

\(\displaystyle t=2sin^2x, dt=4sinxcosxdx ; dx=dt/\sqrt{8t-4t^2}\) Let's try a different substitution

First, have you proved the convergence of the improper integral? I had to look up how to do it. Consider

\(\displaystyle \displaystyle \lim_{h \to 0} \dfrac{h^\alpha}{\sqrt{1 - 2\ \sin^2(\pi/4 - h)}} \)

If you can find \(\displaystyle \alpha < 1\) for which this limit exists, then the integral does converge. Because the sin^2 is multiplied by 2, this integral will NOT converge if the upper limit is greater than pi/4. It looks like an elliptic integral of the first kind, but is not because \(\displaystyle k^2 < 1\) is not satisfied.

In general (?) the substitution t = sinx will switch between trigonometric and power forms:
\(\displaystyle t = \sin x\),......\(\displaystyle dt = \cos x\ dx\),......\(\displaystyle dx = \dfrac{dt}{\sqrt{1 - t^2}}\)

\(\displaystyle \displaystyle \int_0^{1/\sqrt{2}} \dfrac{dt}{\sqrt{1 - t^2}\ \sqrt{1 - 2t^2}} \)

Again, that looks like an elliptical integral of the first kind, EXCEPT \(\displaystyle k^2 = 2 > 1\). Can you make the factors in the denominator look like \(\displaystyle \Gamma(p)\Gamma(q)/\Gamma(p+q)\) ?
Or anything else related to Euler functions?

Good Luck!