Why Does Integration of Certain Functions Result in Infinity?

  • Thread starter Thread starter touqra
  • Start date Start date
  • Tags Tags
    Integration
touqra
Messages
284
Reaction score
0
Using Maple 8, I integrate this:

\int_0^1\frac{1}{\sqrt(p(1-p))} dp

and I get \pi
but, this function goes to infinity at p = 0 and 1.
How can this be possible ?
 
Physics news on Phys.org
How can this be possible ?
Why wouldn't it be?
 
Hurkyl said:
Why wouldn't it be?

The denominator has sqrt[p*(1-p)], which will give infinity on both p = 0 and 1. And so, the area under this curve is infinity too, right ?
 
Can you think of a reason why that should be true, though? If you're convinced the area should be infinite, then try to work out a proof of it -- such exercises are often really good at clearing up misunderstandings.

Incidentally, it might help to play with simpler functions, e.g.

\int_0^1 \frac{1}{\sqrt{x}} \, dx

or

\int_1^\infty \frac{1}{y^2} \, dy
 
Back
Top