Solving a Tricky Integral: Tips and Techniques

[heinerL]

Hey

hope anybody can help me with a tricky integral ( i should check if it exists):

\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{sin(x)}*cos^2(x)} dx


And i have really no idea where to start!

thanks

 

[Dick]

You should definitely check if it exists before you work too hard on it. What's the behavior like at x=0 and x=pi/2?

 

[LCKurtz]

You aren't going to be able to find a "simple" antiderivative. Maple gives a complicated answer using Elliptic functions.

 

[heinerL]

So you mean, because it does not exist for Pi/2 it does not exist at all? so simple?

And yeah, i know i did check it with maple too!

 

[vela]

No, it's not that simple. You have to look at the behavior of the function around the singularities at x=0 and x=π/2. The integral can still converge is the function doesn't blow up too quickly around those two points.

 

[ideasrule]

So you mean, because it does not exist for Pi/2 it does not exist at all? so simple?

That's not necessarily true. Consider the integral of 1/sqrt(x) from x=0 to 1. The integral exists, even though 1/sqrt(x) itself does not exist at x=0.

 

[heinerL]

But if I try the limit x->Pi/2 i still get inf?

 

[Dick]

But if I try the limit x->Pi/2 i still get inf?

The easiest way to see the problem at pi/2 is to expand your integrand in a Taylor series around x=pi/2 and realize the integrand looks like 1/y^2 where y=pi/2-x.

 

[heinerL]

so you mean because the taylor series at x=pi/2 --> 1/(pi/2-x)^2 which is 1/0 at pi/2 and therefore not defined the integral does not exist? if yes, can you please tell me why? thx

 

[Dick]

As others have already said, it's not that the form is 1/0. It's that the integral of 1/y^2 in an interval including 0 doesn't not exist. Take integral from 0 to 1 of 1/y^2 and set in up as an improper integral, i.e. take integral from epsilon to 1 and let epsilon approach 0. Do you get a limit? Now do the same thing with 1/sqrt(y). Do you get a limit. They are both 1/0 AT y=0.