vilhelm
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Would be very grateful if someone helped me on this.
$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
oooppp2 said:Would be very greatful if someone helped me on this.
$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
oooppp2 said:Would be very greatful if someone helped me on this.
$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
ZaidAlyafey said:It can be solved by contour integration or by Beta function .
$$ \int_0^\infty \frac{dx}{1+x^3} =\frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3}-1}}{1+x } dx =\frac{1}{3} \Gamma \left( \frac{1}{3}\right) \Gamma \left(1- \frac{1}{3}\right) = \frac{\pi }{3 \sin \left (\frac{\pi }{3} \right) } = \frac{2\pi }{3 \sqrt{3}}$$
I like Serena said:$x^3+1$ has the root $x=-1$.
Therefore we can factorize it:
$$x^3 + 1 = (x+1)(x^2-x+1)$$
Next step is partial fractions...
oooppp2 said:First calculus course, so I have not learned that.
oooppp2 said:
I like Serena said:Looks like you know how to do it.
Good. ;)
However, in your last 2 lines you write:
\begin{array}{lcl}
1 &=& \frac 1 3 + 2B + \frac 4 3 \\
B &=& -\frac 4 3
\end{array}
But if I substitute that, I get:
\begin{array}{lcl}
1 &=& \frac 1 3 + 2 \cdot (- \frac 4 3) + \frac 4 3 \\
1 &=& -\frac 3 3
\end{array}
But... that does not seem right!For the next steps, you might work toward the following integrals:
\begin{array}{lcl}
\int \frac {du}{1+u^2} &=& \arctan u + C \\
\int \frac {u du}{1+u^2} &=& \frac 1 2 \ln(1+u^2) + C
\end{array}
oooppp2 said:Here is my attempt. I'm still a bit stuck...
https://www.dropbox.com/s/9gub6zq02b5235b/Screenshot 2013-10-26 21.13.45.png
oooppp2 said:Here is my attempt. I'm still a bit stuck...
https://www.dropbox.com/s/9gub6zq02b5235b/Screenshot 2013-10-26 21.13.45.png
I like Serena said:That's a screenshot with no relevant information...
oooppp2 said:Sorry, wrong image. Here is the correct one.
https://www.dropbox.com/s/bghijxpzobb1f5o/2013-10-27 20.26.56.jpg
I like Serena said:For the next steps, you might work toward the following integrals:
\begin{array}{lcl}
\int \frac {du}{1+u^2} &=& \arctan u + C \\
\int \frac {u du}{1+u^2} &=& \frac 1 2 \ln(1+u^2) + C
\end{array}
As an aside, in light of Zaid's answer to this Q, you can use his reside / Beta Integral derivation to consider certain series evaluations...oooppp2 said:Would be very grateful if someone helped me on this.
$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$