Higher Level Integration Marathon & Questions (1 Viewer)

Paradoxica · Feb 20, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
$$ Find : $ \int_0^1{x^x}dx$

$-\sum_{n=1}^{\infty} (-n)^{-n}$

Well, I mean, Sophomore's Dream is nice and all, but there isn't a closed form...

Drsoccerball · Feb 20, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$-\sum_{n=1}^{\infty} (-n)^{-n}$

Well, I mean, Sophomore's Dream is nice and all, but there isn't a closed form...

Yes that's the answer how did you do it though

? Did you create a taylor series and just integrate each term or?

Paradoxica · Feb 20, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
Yes that's the answer how did you do it though ? Did you create a taylor series and just integrate each term or?

Inspection

InteGrand · Feb 20, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
Yes that's the answer how did you do it though ? Did you create a taylor series and just integrate each term or?

Yeah, it can be proved that way: https://en.wikipedia.org/wiki/Sophomore's_dream#Proof .

Paradoxica · Feb 20, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$\int_0^\infty \frac{(x+1)\ln^2{(x+1)}}{(4x^2 +8x +5)^{\frac{3}{2}}}\,$d$x$

I can provide a hint if anyone is attempting this.

Paradoxica · Feb 20, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$$\noindent Yes, $\frac{5\pi^2}{96}$ is the correct answer but surely you didn't just guess it?$

I have not found a complete solution to this integral, only generalised forms. Do you happen to have a solution that specifically deals with this integral?

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$Now, this is a wild guess, but is the second one $\frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$

Paradoxica said:
Inspection

And by 'inspection' you mean 'Google' or 'Wolfram'?

Nobody's buying it, mate.

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
And by 'inspection' you mean 'Google' or 'Wolfram'?

Nobody's buying it, mate.

It was a joke -_-

I've already encountered Sophomore's Dream before, on one of my many followed blogs.

As for Ahmed's Integral, I didn't know the answer was exactly that until I put the numerical answer into ISC+.

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
It was a joke -_-

I've already encountered Sophomore's Dream before, on one of my many followed blogs.

Didn't look like a joke to me. Looked like taking credit where credit was not due.

But I'll humour you. Explain the your first "wild guess" then.

I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
Didn't look like a joke to me. Looked like taking credit where credit was not due.

But I'll humour you. Explain the your first "wild guess" then.

I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.

From above: As for Ahmed's Integral, I didn't know the answer was exactly that until I put the numerical answer into ISC+.

Also, putting the integral into Wolfram Alpha doesn't give you the closed form. Google... well you can't exactly search tex code easily. So I resorted to the above, taking the numerical value of the integral to 50 decimal places.

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.

Well I would post my solution which I figured out yesterday, but I'm not sure it's worth anything at this point.

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Well I would post my solution which I figured out yesterday, but I'm not sure it's worth anything at this point.

No, I must insist.

Just a brief and general outline will do.

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
Just a brief and general outline will do.

$$\noindent Firstly, use the fact that the domain of integration is strictly on the positive axis to invoke $\tan^{-1}{x} \equiv \frac{\pi}{2} - \tan^{-1}{\frac{1}{x}} \\ \int_0^1 \frac{\tan^{-1}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}} \,$d$x = \frac{\pi}{2} \int_0^1 \frac{\text{d}x}{(x^2+1) \sqrt{2+x^2}}-\int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+x^2}}}}{x^2+1}\\$The first integral evaluates to $\frac{\pi^2}{12}$, which I will leave for the reader to verify. The second integral is not as straightforward. However, observe that the numerator is extremely similar to a standard form from integration tables $\frac{1}{a}\tan^{-1}{\frac{1}{a}} = \int_0^1 \frac{\text{d}y}{y^2+a^2} \\$Here, we borrow the concept of iterated integrals and double integrals from the Gaussian Integral commonly seen in the harder questions of high school.$ \\ \int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+x^2}}}}{x^2+1} = \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+x^2+2)}$

$$\noindent Observe that $y^2+x^2+2 - (x^2+1) = y^2+1$ The integrand can then be decomposed through the method of partial fractions to obtain $\\\int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+1)} - \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(y^2+1)(y^2+x^2+2)}\\$By the principle of bound and unbound variables for definite integrals, the integral is equal to the simpler looking integral minus itself. Thus, we have $\\2\int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+x^2+2)} = \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+1)}\\ $This is now a double integral over the Cartesian plane with identical bounds, and so we have $2\int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+x^2}}}}{x^2+1} = \left ( \int_0^1 \frac{\text{d}r}{r^2+1} \right )^2 = \frac{\pi^2}{16} \\ $And so we conclude $\int_0^1 \frac{\tan^{-1}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}} \,$d$x = \frac{\pi^2}{12}-\frac{\pi^2}{32} = \frac{5\pi^2}{96}$

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Thank you. The spirit of this thread exists in the solution methods, not the answers.

This is still not fully consistent with the below post, but I'll drop the case as an answer has now been given, regardless of its true origins.

Paradoxica said:
I have not found a complete solution to this integral, only generalised forms. Do you happen to have a solution that specifically deals with this integral?

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
Thank you. The spirit of this thread exists in the solution methods, not the answers.

This is still not fully consistent with the below post, but I'll drop the case as an answer has now been given, regardless of its true origins.

In saying so, I can't do anything that can only be done through contour integration, so my box of tools is much smaller than yours, or most people who are on this thread.

Drsoccerball · Feb 21, 2016

Re: Extracurricular Integration Marathon

How do double integrals even work ?

leehuan · Feb 21, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
How do double integrals even work ?

You just evaluate them variable by variable.

How you generate them is a more interesting story.

Paradoxica · Feb 21, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
How do double integrals even work ?

leehuan said:
You just evaluate them variable by variable.

How you generate them is a more interesting story.

Yeah, knowing when to revert a function into the definite integral of a simpler function is not at all obvious for the most part.

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
How do double integrals even work ?

You're probably clever enough to get the general gist of it (double integration via Cartesian coordinate system) from this image. I'm guessing you just want a rough idea, not a more rigorous construction.

Just imagine throwing in an extra dimension to everything. Instead of 1 dimensional partitions (the subintervals) of a domain, we have 2 dimensional partitions of a region. Instead of taking 2 dimensional 'strips', we have 3 dimensional 'strips'. Instead of swooping over the domain once, we swoop over it twice (once in the x direction, again in the y direction). Instead of finding an 'area', we find a 'volume'.

Drsoccerball · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
You're probably clever enough to get the general gist of it (double integration via Cartesian coordinate system) from this image. I'm guessing you just want a rough idea, not a more rigorous construction.

Just imagine throwing in an extra dimension to everything. Instead of 1 dimensional partitions (the subintervals) of a domain, we have 2 dimensional partitions of a region. Instead of taking 2 dimensional 'strips', we have 3 dimensional 'strips'. Instead of swooping over the domain once, we swoop over it twice (once in the x direction, again in the y direction). Instead of finding an 'area', we find a 'volume'.

Is this apart of Matrices ? Doesn't seem to hard

So can we treat one of the changing functions as a constant while we integrate the 'flat' part ?

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