Welcome to our community

Be a part of something great, join today!

Integrate dx/(1+x^3)

oooppp2

New member
Oct 21, 2013
19
Would be very grateful if someone helped me on this.

$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Would be very greatful if someone helped me on this.

$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
$x^3+1$ has the root $x=-1$.
Therefore we can factorize it:
$$x^3 + 1 = (x+1)(x^2-x+1)$$
Next step is partial fractions...
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Would be very greatful if someone helped me on this.

$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
It can be solved by contour integration or by Beta function .

\(\displaystyle \int_0^\infty \frac{dx}{1+x^3} =\frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3}-1}}{1+x } dx =\frac{1}{3} \Gamma \left( \frac{1}{3}\right) \Gamma \left(1- \frac{1}{3}\right) = \frac{\pi }{3 \sin \left (\frac{\pi }{3} \right) } = \frac{2\pi }{3 \sqrt{3}}\)
 

oooppp2

New member
Oct 21, 2013
19
It can be solved by contour integration or by Beta function .

\(\displaystyle \int_0^\infty \frac{dx}{1+x^3} =\frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3}-1}}{1+x } dx =\frac{1}{3} \Gamma \left( \frac{1}{3}\right) \Gamma \left(1- \frac{1}{3}\right) = \frac{\pi }{3 \sin \left (\frac{\pi }{3} \right) } = \frac{2\pi }{3 \sqrt{3}}\)
First calculus course, so I have not learned that.

- - - Updated - - -

$x^3+1$ has the root $x=-1$.
Therefore we can factorize it:
$$x^3 + 1 = (x+1)(x^2-x+1)$$
Next step is partial fractions...
Thanks, I'll try it and update with my progress.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
First calculus course, so I have not learned that.
I hope I didn't create confusion for you. It is important to keep in mind that this problem can be solved by complex analysis approaches , if you are going to take this course in the future .
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
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

New member
Oct 21, 2013
19
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}
Here is my attempt. I'm still a bit stuck...

https://www.dropbox.com/s/9gub6zq02b5235b/Screenshot 2013-10-26 21.13.45.png
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This link looks to me like a screen shot taken while you were composing a post.

Let's look at the partial fraction decomposition:

\(\displaystyle \frac{1}{x^3+1}=\frac{1}{(x+1)\left(x^2-x+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\)

Multiplying through by \(\displaystyle (x+1)\left(x^2-x+1 \right)\) we obtain:

\(\displaystyle 1=A\left(x^2-x+1 \right)+(Bx+C)(x+1)\)

\(\displaystyle 1=Ax^2-Ax+A+Bx^2+Bx+Cx+C\)

\(\displaystyle 0x^2+0x+1=(A+B)x^2+(-A+B+C)x+(A+C)\)

Comparison of coefficients results in the linear system:

\(\displaystyle A+B=0\)

\(\displaystyle -A+B+C=0\)

\(\displaystyle A+C=1\)

What do you find when you solve this system?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779

oooppp2

New member
Oct 21, 2013
19
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}
But for the arctan(u) and ln(1+u^2) should I take the limit as n goes to infty? How is that done? Thanks.
 

oooppp2

New member
Oct 21, 2013
19

DreamWeaver

Well-known member
Sep 16, 2013
337
Would be very grateful if someone helped me on this.

$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$


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...

Let

\(\displaystyle I_n=\int_0^{\infty}\frac{dx}{(1+x^n)}\)


Specifically, it might be interesting to expand the \(\displaystyle (1+x^n)\) term in the denominator via

\(\displaystyle (1+x^n)=\sum_{k=0}^{\infty}(-1)^kx^{kn}\)

However, this series has radius of convergence \(\displaystyle r=1\), so first we'd need to split the integral into:

\(\displaystyle I_n=\int_0^1 \frac{dx}{(1+x^n)}+\int_1^{\infty}\frac{dx}{(1+x^n)}\)


Now apply the substitution \(\displaystyle x=1/y\), \(\displaystyle dx=-dy/y^2\) on the second integral to get:


\(\displaystyle I_n=\int_0^1\frac{dx}{(1+x^n)}-\int_1^0\frac{dy}{y^2(1+(1/y)^n)}=\)


\(\displaystyle \int_0^1 \Bigg\{ \frac{1}{1+x^n}+\frac{x^{n-2}}{(1+x^n)} \Bigg\} \,dx= \int_0^1\frac{1}{1+x^n}(1+x^{n-2})\,dx=\)


\(\displaystyle \sum_{k=0}^{\infty} (-1)^k\int_0^1x^{kn}(1+x^{n-2})\,dx\)



From a historical point of view, this type of integral is particularly interesting with regards to the infamous calculus battle between Leibniz and Newton... Leibniz used the case n=2 to prove what is now famously, and indeed mistakenly (!!!), called Leibniz's Series:

\(\displaystyle \int_0^1\frac{dx}{1+x^2}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7} + \cdots = \frac{\pi}{4}
\)

[This series was known to Madhava in the Fourteenth Centurty :eek: )


To which Newton replied by evaluating the case for n=4 in two different ways, to show that:

\(\displaystyle \int_0^1\frac{1+x^2}{1+x^4}\,dx=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}- \cdots = \frac{\pi}{2 \sqrt{2}}\)




Ermmm... Seem to have wandered a tad off-topic here. Whoops! (Heidy)


Back on-topic: If you consider Zaid's answer above, and the series form given here, then you have two different evaluations of each integral - much like Sir Isaac Newton - for the parameter n. Standing on the shoulders of giants indeed... (Rock)