# Integrate dx/(1+x^3)

#### oooppp2

##### New member
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
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
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
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
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
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
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

Staff member

#### MarkFL

Staff member
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?

Staff member

#### oooppp2

##### New member
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.

#### DreamWeaver

##### Well-known member
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 )

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!

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...