Can you evaluate this integral using series and historical methods?

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In summary, the integral $\displaystyle\int_0^\infty \frac{dx}{1+x^3}$ can be solved using contour integration or the Beta function. It can also be approached by considering a series expansion of the denominator and splitting the integral into two parts. This method has historical significance in the infamous calculus battle between Leibniz and Newton.
  • #1
vilhelm
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Would be very grateful if someone helped me on this.

$\displaystyle\int_0^\infty \frac{dx}{1+x^3}$
 
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  • #2
oooppp2 said:
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...
 
  • #3
oooppp2 said:
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 .

\(\displaystyle \int_0^\infty \frac{dx}{1+x^3} =\frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3}-1}}{1+x } dx =\frac{1}{3} \Gamma \left( \frac{1}{3}\right) \Gamma \left(1- \frac{1}{3}\right) = \frac{\pi }{3 \sin \left (\frac{\pi }{3} \right) } = \frac{2\pi }{3 \sqrt{3}}\)
 
  • #4
ZaidAlyafey said:
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 - - -

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...

Thanks, I'll try it and update with my progress.
 
  • #5
oooppp2 said:
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 .
 
  • #7
oooppp2 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}
 
  • #8
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}

Here is my attempt. I'm still a bit stuck...

https://www.dropbox.com/s/9gub6zq02b5235b/Screenshot 2013-10-26 21.13.45.png
 
  • #10
oooppp2 said:

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?
 
  • #13
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}

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.
 
  • #15
oooppp2 said:
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)
 

FAQ: Can you evaluate this integral using series and historical methods?

What is the general formula for integrating dx/(1+x^3)?

The general formula for integrating dx/(1+x^3) is (1/3)ln|1+x^3| + C, where C is the constant of integration.

Can I use u-substitution to solve this integral?

Yes, you can use u-substitution by letting u = 1 + x^3 and du = 3x^2 dx. This will result in the integral becoming (1/3)∫du/u, which can be easily integrated using the natural logarithm function.

Is there a way to solve this integral without using u-substitution?

Yes, you can also use partial fractions to solve this integral. By decomposing dx/(1+x^3) into (A+Bx)/(1+x^3), where A and B are constants, you can then solve for A and B and integrate each term separately.

Can I use trigonometric substitution to solve this integral?

Yes, you can use trigonometric substitution by letting x = tanθ and using the identity 1 + tan^2θ = sec^2θ. This will result in the integral becoming (1/3)∫secθdθ, which can be easily solved using the natural logarithm function.

Are there any real-world applications of this integral?

Yes, this integral can be used in calculus and physics to calculate the area under a curve. It can also be used in engineering and economics to solve problems involving optimization and rates of change.

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