integral from 0 to 1: (1 - x^7)^(1/3)...
- Thread starter dts5044
- Start date
I got to looking at your problem and noticed it looks like we can use the Beta function.
\(\displaystyle B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\)
We have to use a little substitution to hammer into shape.
For the first one:
\(\displaystyle \int_{0}^{1}(1-x^{7})^{\frac{1}{3}}dx\)
Let \(\displaystyle u=x^{7}, \;\ u^{\frac{1}{7}}=x, \;\ \frac{1}{7}u^{\frac{-6}{7}}du=dx\)
\(\displaystyle \frac{1}{7}\int_{0}^{1}u^{\frac{-6}{7}}(1-u)^{\frac{1}{3}}du\)
Now, we have \(\displaystyle p-1=\frac{-6}{7}, \;\ p=\frac{1}{7}\)
\(\displaystyle q-1=\frac{1}{3}, \;\ q=\frac{4}{3}\)
Now, sub into our gamma functions:
\(\displaystyle \boxed{\frac{1}{7}\cdot\frac{\Gamma(\frac{1}{7})\cdot\Gamma(\frac{4}{3})}{\Gamma(\frac{1}{7}+\frac{4}{3})}=.943132664565}\)
If you use the same technique on the other one, you find the same exact answer. Subtracting them yields 0.
\(\displaystyle B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\)
We have to use a little substitution to hammer into shape.
For the first one:
\(\displaystyle \int_{0}^{1}(1-x^{7})^{\frac{1}{3}}dx\)
Let \(\displaystyle u=x^{7}, \;\ u^{\frac{1}{7}}=x, \;\ \frac{1}{7}u^{\frac{-6}{7}}du=dx\)
\(\displaystyle \frac{1}{7}\int_{0}^{1}u^{\frac{-6}{7}}(1-u)^{\frac{1}{3}}du\)
Now, we have \(\displaystyle p-1=\frac{-6}{7}, \;\ p=\frac{1}{7}\)
\(\displaystyle q-1=\frac{1}{3}, \;\ q=\frac{4}{3}\)
Now, sub into our gamma functions:
\(\displaystyle \boxed{\frac{1}{7}\cdot\frac{\Gamma(\frac{1}{7})\cdot\Gamma(\frac{4}{3})}{\Gamma(\frac{1}{7}+\frac{4}{3})}=.943132664565}\)
If you use the same technique on the other one, you find the same exact answer. Subtracting them yields 0.
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
- Messages
- 108
Find the power series for (1-x^7)^(1/3). Use binomial series.
[1-x^7]^(1/3) = 1-x^7/3-x^14/9-5x^21/81-10x^28/243-22x^35/729-154x^42/6561+...+
Integral[1-x^7]^(1/3),x,0,1 = x-x^8/24-x^15/135-5x^22/1782-10x^29/2047-22x^36/26244-154x^43/282123-...,x,0,1
+1-1/24-1/135-5/1782-10/7047-22/26244-154/282123 = about .94531 which with only 7 terms the value is only off .00218. Good enough for government work.
[1-x^7]^(1/3) = 1-x^7/3-x^14/9-5x^21/81-10x^28/243-22x^35/729-154x^42/6561+...+
Integral[1-x^7]^(1/3),x,0,1 = x-x^8/24-x^15/135-5x^22/1782-10x^29/2047-22x^36/26244-154x^43/282123-...,x,0,1
+1-1/24-1/135-5/1782-10/7047-22/26244-154/282123 = about .94531 which with only 7 terms the value is only off .00218. Good enough for government work.