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- Mar 5, 2012

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$x^3+1$ has the root $x=-1$.Would be very greatful if someone helped me on this.

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

Therefore we can factorize it:

$$x^3 + 1 = (x+1)(x^2-x+1)$$

Next step is partial fractions...

- Jan 17, 2013

- 1,667

It can be solved by contour integration or by Beta function .Would be very greatful if someone helped me on this.

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

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

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First calculus course, so I have not learned that.It can be solved by contour integration or by Beta function .

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

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Thanks, I'll try it and update with my progress.$x^3+1$ has the root $x=-1$.

Therefore we can factorize it:

$$x^3 + 1 = (x+1)(x^2-x+1)$$

Next step is partial fractions...

- Jan 17, 2013

- 1,667

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 .First calculus course, so I have not learned that.

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- Mar 5, 2012

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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}

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

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}

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

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- Mar 5, 2012

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That's a screenshot with no relevant information...Here is my attempt. I'm still a bit stuck...

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

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This link looks to me like a screen shot taken while you were composing a post.Here is my attempt. I'm still a bit stuck...

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

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?

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Sorry, wrong image. Here is the correct one.That's a screenshot with no relevant information...

https://www.dropbox.com/s/bghijxpzobb1f5o/2013-10-27 20.26.56.jpg

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- Mar 5, 2012

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Oh, oh.Sorry, wrong image. Here is the correct one.

https://www.dropbox.com/s/bghijxpzobb1f5o/2013-10-27 20.26.56.jpg

For some reason you wrote $x^2 - x - 1$ right at the beginning while it should be $x^2 - x + 1$.

Your previous attempt was pretty good really, you only made a small mistake at the end.

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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.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}

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I try to find primitve functions. Wolfram Alpha primitive \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1} - Wolfram|Alpha

and I disagree

My answer is this

4/3arctan(?4\?3*(x-1/2))+1/2(ln(x^2-x+1)+4/3arctan(?4/?3(x-1/2 - Wolfram|Alpha)))

And here are my calculations

https://www.dropbox.com/sh/5a76xairfirt39b/S7XE0q_Exh

- 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 )

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!