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## Integration Contest - Season 4

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### Re: Integration Contest - Season 4

Tue Mar 11, 2014 8:39 am
Random Variable Integration Guru

Posts: 381
Let $\displaystyle f(z) = \frac{\sqrt[3]{z^{2}(z-1)}}{(1+z)^{3}}$ where $\displaystyle 0 \le \arg(z), \arg(z-1) \le 2 \pi$.

Then integrating around a dumbbell contour,

$\displaystyle 2 \pi i a_{-1} + \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x) e^{i \pi}}}{(1+x)^{3}} \ dx + \int_{1}^{0} \frac{\sqrt[3]{(xe^{2 \pi i})^{2} (1-x) e^{i \pi}}}{(1+x)^{3}} \ dx = 2 \pi i \text{Res} [f(z),-1]$

where $a_{-1}$ is the coefficient of the $\frac{1}{z}$ term in the Laurent expansion at infinity, that is, $\displaystyle -\text{Res}[f(z), \infty]$.

$\displaystyle 0 + 2 i \sin \left( \frac{\pi}{3} \right) \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+x)^{3}} \ dx = - \frac{2 \pi i }{9} \lim_{z \to -1} x^{2} \Big( x^{2}(x-1) \Big)^{-5/3} = - \frac{2 \pi i}{9} \Big( (e^{\pi i})^{2} 2 e^{\pi i} \Big)^{-5/3}$

$\displaystyle = - \frac{2 \pi i}{9} 2^{-5/3} e^{-5 \pi i} = \frac{\pi i}{9} 2^{-2/3}$

Which implies

$\displaystyle \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+x)^{3}} \ dx = \frac{\pi \ 2^{-2/3}}{9 \sqrt{3}} = \frac{ \pi \sqrt[3]{2}}{18 \sqrt{3}}$

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 9:16 am

Posts: 852
Location: Jaipur, India

Very nice RV!

Actually there is a more generalized formula for these kind of integrals.

$\displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \frac{dx}{(x+p)^{a+b}} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{1}{(1+p)^{a} p^b}$

http://integralsandseries.prophpbb.com/post72.html#p72

Put $a=\frac{5}{3},\; b=\frac{4}{3}$ and $p=1$:

\displaystyle \begin{align*} \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+x)^{3}} \ dx &= \frac{\Gamma\left( \frac{5}{3}\right)\Gamma\left( \frac{4}{3}\right)}{2^{\frac{5}{3}}\cdot 2!}\\ &= \frac{\pi}{9\cdot 2^{\frac{5}{3}}\sin\left(\frac{\pi}{3} \right)} \\ &= \frac{\pi \sqrt[3]{2}}{18\sqrt{3}} \end{align*}

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 10:31 am
Random Variable Integration Guru

Posts: 381
Cody probably didn't intend for the integral to be of that specific form.

$\displaystyle \int_{0}^{1} \frac{x^{a-1} (1-x)^{b-1}}{(x+p)^{a+b}} \ dx = \frac{1}{p^{a+b}} \int_{0}^{1} x^{a-1} (1-x)^{b-1} \left( 1 + \frac{x}{p} \right)^{-(a+b)} \ dx$

$\displaystyle = \frac{B(a,b)}{p^{a+b}} \ _{2}F_{1} \left(a+b, a; a+b; -\frac{1}{p} \right) = \frac{B(a,b)}{p^{a+b}} \ _{1}F_{0} \left( a; - ; -\frac{1}{p} \right)$

$\displaystyle = \frac{B(a,b)}{p^{a+b}} \left( 1+ \frac{1}{p} \right)^{-a}= \frac{B(a,b)}{p^{b} (1+p)^{a}}$

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 10:35 am

Posts: 852
Location: Jaipur, India

Has anybody seen a proof for the following identity?

Problem 35 (Optional)

$\displaystyle \int_0^1 \frac{x^{\mu+1/2}(1-x)^{\mu-1/2}}{(c+2bx-ax^2)^{\mu+1}}dx = \frac{\beta \left( \frac{1}{2},\mu+\frac{1}{2}\right)}{\left\{a+(\sqrt{c}+\sqrt{c+2b-a})^2\right\}^{\mu+1/2}\sqrt{c+2b-a}}$

$\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the Beta Function.

This equation is highly generalized and a large variety of integrals can be solved with it's help.

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 10:47 am

Posts: 852
Location: Jaipur, India

And here's another:

Problem 36

$\displaystyle \int_0^\infty F(z)\,F\left(z\,\sqrt2\right)\frac{e^{-z^2}}{z^2}dz=\frac{\pi^{3/2}}8\left(\sqrt2-4\right)+\frac{3\,\pi^{1/2}}2\tan^{-1}\sqrt2$

where $\displaystyle F(x)=e^{-x^2}\int_0^x e^{y^2}\; dy$

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 3:48 pm
galactus
Global Moderator

Posts: 902
Random Variable wrote:
Let $\displaystyle f(z) = \frac{\sqrt[3]{z^{2}(z-1)}}{(1+z)^{3}}$ where $\displaystyle 0 \le \arg(z), \arg(z-1) \le 2 \pi$.

Then integrating around a dumbbell contour,

$\displaystyle 2 \pi i a_{-1} + \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x) e^{i \pi}}}{(1+z)^{3}} \ dx + \int_{1}^{0} \frac{\sqrt[3]{(xe^{2 \pi i})^{2} (1-x) e^{i \pi}}}{(1+z)^{3}} \ dx = 2 \pi i \text{Res} [f(z),-1]$

where $a_{-1}$ is the coefficient of the $\frac{1}{z}$ term in the Laurent expansion at infinity, that is, $\displaystyle -\text{Res}[f(z), \infty]$.

$\displaystyle 0 + 2 i \sin \left( \frac{\pi}{3} \right) \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+z)^{3}} \ dx = - \frac{2 \pi i }{9} \lim_{z \to -1} x^{2} \Big( x^{2}(x-1) \Big)^{-5/3} = - \frac{2 \pi i}{9} \Big( (e^{\pi i})^{2} 2 e^{\pi i} \Big)^{-5/3}$

$\displaystyle = - \frac{2 \pi i}{9} 2^{-5/3} e^{-5 \pi i} = \frac{\pi i}{9} 2^{-2/3}$

Which implies

$\displaystyle \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+x)^{3}} \ dx = \frac{\pi \ 2^{-2/3}}{9 \sqrt{3}} = \frac{ \pi \sqrt[3]{2}}{18 \sqrt{3}}$

RV, buddy, may I ask you something?. Or, whomever wants to add to the dialogue. It's driving me nuts. I am a little fuzzy about how to do these types of integrals when one uses contours. In other words, determining the arguments with a dogbone. How in the world do you determine the arguments?. I have noticed they are different depending on the problem.

Sometimes you use $\displaystyle -\pi<arg(1-z)\leq \pi$ and other times $\displaystyle 0<arg(1-z)\leq 2\pi$ or something like that. My inequalities may not be exact.

Take the last one you just done. You used $\displaystyle e^{\pi i}$ inside the radical and said the argument for z and 1-z are between 0 and 2pi. Why not make the argument for z between -pi and pi... as it is many times I have seen?.

Take this one:

$\displaystyle \int_{0}^{1}\frac{\sqrt{x(1-x)}}{(1+x^{2})^{2}}=\frac{\pi}{8}\sqrt{\sqrt{2}-1}$

I found the sum of the residues at -i and i to be $\displaystyle \frac{\pi i}{4}\sqrt{\sqrt{2}+1}$

which would mean we would have to find something like $\displaystyle 2(\sqrt{2}+1)i$ in the two integrals above and below the branch so it could be divided into the residue and get the correct result.

But, what arguments would you use for this one?. I have looked at that diagram they have on Wikipedia for 'methods of contour integration', but I do not understand why the circle around 3 is 0 below the branch and 2pi above, and the circle around 0 is pi above and -pi below the branch.

There is something about the argument business with these that has be a little befuddled.

I can make them work, but not sure why we use what we do and when.

For instance, I saw one of your old posts on SE where achillehui said that the argument for $\displaystyle \sqrt{1-x^{2}}$ was $\displaystyle \pi/2$. From where is he getting this?. How do you tell what the argument for something like this is?. I am missing something probably obvious with this stuff, but it is driving me crazy.

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 5:51 pm
Random Variable Integration Guru

Posts: 381
Let me go back to the problem I did above.

Let $\displaystyle f(z) = \frac{z^{2/3} (z-1)^{1/3}}{(1+z)^{3}} = \frac{(|z|e^{i \arg z})^{2/3} (|z-1|e^{i \arg(z-1)})^{1/3} }{(1+z)^{3}}$.

By choosing the branch cut $\displaystyle [0, \infty)$ for $\displaystyle z^{2/3}$ and the branch cut $\displaystyle [1, \infty)$ for $\displaystyle (z-1)^{1/3}$, we in essence have a single cut along $\displaystyle [0, \infty)$.

But just above the real axis and to the right of $\displaystyle z=1$, $\displaystyle \arg(z) = \arg(z-1) = 0$.

And just below the real axis and to the right of $\displaystyle z=1$, $\displaystyle \arg(z) = \arg(z-1) = 2 \pi$.

So just above the real axis and to the right of $\displaystyle z=1$, $\displaystyle f(z) = \frac{(xe^{0i})^{2/3} ((x-1)e^{0i})^{1/3} }{(1+x)^{3}} = \frac{x^{2/3} (x-1)^{1/3}}{(1+x)^{3}}$.

And just below the real axis and to the right of $\displaystyle z = 1$, $\displaystyle f(z) = \frac{(x e^{2 \pi i})^{2/3} ((x-1)e^{2 \pi i})^{1/3} }{(1+x)^{3}} = e^{2 \pi i} \frac{x^{2/3} (x-1)^{1/3}}{(1+x)^{3}} = \frac{x^{2/3} (x-1)^{1/3}}{(1+x)^{3}}$.

So no cut is needed on $\displaystyle [1, \infty)$ and a simply a cut on $\displaystyle [0,1]$ will suffice.

Next we determine the value of $\displaystyle f(z)$ just above and just below the cut on $\displaystyle [0,1]$.

Just above the cut $\displaystyle \arg(z) = 0$ and $\displaystyle \arg(z-1) = \pi$.

And just below the cut $\displaystyle \arg(z) = 2 \pi$ and $\displaystyle \arg (z-1) = \pi$.

So just above the cut $\displaystyle f(z) = \frac{(xe^{i0})^{2/3} (|x-1|e^{\pi i})^{1/3} }{(1+x)^{3}} = e^{\pi i /3} \frac{x^{2/3} (1-x)^{1/3}}{(1+x)^{3}}$

And just below the cut $\displaystyle f(z) = \frac{(xe^{2 \pi i })^{2/3} (|x-1|e^{\pi i})^{1/3} }{(1+z)^{3}} = e^{- \pi i /3} \frac{x^{2/3} (1-x)^{1/3}}{(1+x)^{3}}$

I always go clockwise around the dumbbell so that I go counterclockwise around $|z|=R$.

Also if there is any ambiguity about the residue at infinity, we know that it must be real-valued along the real axis.

Notice that I used $\displaystyle f(z) = \frac{z^{2/3} (z-1)^{1/3}}{(1+z)^{3}}$, and not $\displaystyle f(z) = \frac{z^{2/3} (1-z)^{1/3}}{(1+z)^{3}}$.

If I choose the later, I would need to do what is done in that Wikipedia article. It's a bit more confusing to deal with $\displaystyle (1-z)^{a}$ than $(z-1)^{a}$.

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 6:26 pm
galactus
Global Moderator

Posts: 902
Thanks RV for taking the time to type that up. I understand what you done there. I appreciate your time. Especially when I'm being thick.

My issue lies in picturing why:

"Next we determine the value of $\displaystyle f(z)$ just above and just below the cut on $\displaystyle [0,1]$.

Just above the cut $\displaystyle \arg(z) = 0$ and $\displaystyle \arg(z-1) = \pi$.

And just below the cut $\displaystyle \arg(z) = 2 \pi$ and $\displaystyle \arg (z-1) = \pi$."

EDIT:

I think I am beginning to see the idea. I drew a picture and labeled the vectors with arrows.

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 7:29 pm
Random Variable Integration Guru

Posts: 381
I choose the branch cut $[1, \infty)$ for $(z-1)^{1/3}$ which restricts $\displaystyle \arg (z-1)$ to $\displaystyle \arg(z-1) \in (0, 2 \pi ]$.

So for $(z-1)^{1/3}$ there is no cut to the left of $z=1$, which means the argument of $(z-1)^{1/3}$ is the same just above and just below the real axis to the left of $z=1$.

EDIT: Yes, it makes more sense with a picture.

### Re: Integration Contest - Season 4

Tue Mar 11, 2014 7:39 pm
galactus
Global Moderator

Posts: 902
thanks RV. I recently wanted a thorough understanding of this idea.

I was trying a few integrals this way and realized..oops...I better brush up on it.

i.e. that one I mentioned earlier. $\displaystyle \int_{0}^{1}\frac{\sqrt{x(1-x)}}{(x^{2}+1)^{2}}dx$.

I wanted to try this one that way, but ran into a snafu

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