Why Does an Integrand Equaling Zero at x=1 Not Determine the Integral's Value?

  • I
  • Thread starter Hornbein
  • Start date
  • Tags
    Integral
  • #1
Hornbein
2,239
1,836
I'm trying to calculate the volume of a truncated hypersphere. As part of it I want this integral.
Integral.jpg

hypergeometric.jpg


Clearly when x=1 the integrand is zero. But plugging this into the series give me a number greater than one. It is true that the series is not defined for x=1, but subtracting some tiny sum isn't going to make any difference.

I hope I'm missing something obvious.
 
Physics news on Phys.org
  • #2
I made a mistake in my previous post ( deleted it ).

$$
\begin{align}
\int (1-x^2)^n ~dx &= \int (-x^2+1)^n ~dx \nonumber\\
&=\int \sum_{k=0}^n \binom {n}{k} (-x^2)^k1^{n-k}~dx \nonumber\\
&=\sum_{k=0}^n (-1)^k \binom {n}{k} \int x^{2k}~dx \nonumber\\
&=\sum_{k=0}^n (-1)^k \frac { n!}{k!(n-k)!} \cdot \frac {x^{2k+1}}{2k+1} \nonumber\\
\end{align}
$$
 
Last edited:
  • #3
Hornbein said:
Clearly when x=1 the integrand is zero. But plugging this into the series give me a number greater than one. It is true that the series is not defined for x=1, but subtracting some tiny sum isn't going to make any difference.

I hope I'm missing something obvious.

Why would the integrand being zero at [itex]x = 1[/itex] tell you anything about the value of the integral?
 

Similar threads

Replies
14
Views
2K
  • Calculus
Replies
8
Views
2K
Replies
1
Views
1K
Replies
6
Views
2K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
4
Views
655
Back
Top