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PostPosted: Tue, 30 Dec 2008 10:12:29 UTC 
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Evaluate: $ f(x_n)= \int_0^{x_n} \int_0^{x_{n-1}} \cdots \int_0^{x_1} \ln x\ dx \ dx_1 \cdots \ dx_{n-2} \ dx_{n-1}.


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PostPosted: Tue, 30 Dec 2008 19:52:17 UTC 
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commutative wrote:
Evaluate: $ f(x_n)= \int_0^{x_n} \int_0^{x_{n-1}} \cdots \int_0^{x_1} \ln x\ dx \ dx_1 \cdots \ dx_{n-2} \ dx_{n-1}.


Spoiler:
Easy to prove
\displaystyle
\int_0^x t^{r-1}\log t\,\mathrm{d}t=\frac{1}{r}\left[x^r\log x-\frac{1}{r}x^r\right]
and so it is just a matter of fiddling about... to give
\displaystyle
f(y)=\frac{1}{n!}x^{n}\log x-\frac{H_n}{n!}x^{n}
where H_n=1+(1/2)+(1/3)+...+(1/n) is the n-th Harmonic number.


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PostPosted: Wed, 31 Dec 2008 00:57:41 UTC 
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outermeasure wrote:
commutative wrote:
Evaluate: $ f(x_n)= \int_0^{x_n} \int_0^{x_{n-1}} \cdots \int_0^{x_1} \ln x\ dx \ dx_1 \cdots \ dx_{n-2} \ dx_{n-1}.


Spoiler:
Easy to prove
\displaystyle
\int_0^x t^{r-1}\log t\,\mathrm{d}t=\frac{1}{r}\left[x^r\log x-\frac{1}{r}x^r\right]
and so it is just a matter of fiddling about... to give
\displaystyle
f(y)=\frac{1}{n!}x^{n}\log x-\frac{H_n}{n!}x^{n}
where H_n=1+(1/2)+(1/3)+...+(1/n) is the n-th Harmonic number.


your answer is correct. it can be proved by induction that: $ f(x_n)=\frac{x_n^n}{n!}\left(\ln(x_n) - \sum_{k=1}^n \frac{1}{k} \right).


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PostPosted: Wed, 31 Dec 2008 08:57:40 UTC 
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Oops, it should be f(x) rather than f(y).


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PostPosted: Wed, 31 Dec 2008 19:30:16 UTC 
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Here's an interesting little variation on this I found while doing the Putnam exam (the above problem was about half of the 2008 B2) (sorry if this is hi-jacking, but it's basically the same problem but requires a bit more thinking):
Evaluate $ \int_{0}^{x} \ln(t) (x-t)^{n} dt

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PostPosted: Wed, 31 Dec 2008 22:45:50 UTC 
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cody wrote:
As for your integral, \int_{0}^{x}ln(x)(x-t)^{n}dt

I get:

\frac{x^{n+1}ln(x)}{n+1}-\displaystyle\sum_{k=1}^{n+1}\frac{1}{k}\cdot\frac{x^{n+1}}{n+1}


Yep, that's it. I'm curious if you used the Cauchy formula for repeated integration or did the calculations from scratch?

ps: A really fun one is finding the length of y=1/x in the coordinate system \overline{x} = Re\left[ \sin(x + iy)\right], \overline{y} = Im \left[ \sin(x + iy) \right] (note that it's not an orthogonal coordinate system, i.e. it's metric tensor has off-diagonal entries)

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Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
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PostPosted: Thu, 1 Jan 2009 19:21:52 UTC 
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cody wrote:
I hope no one gets peeved about this, but regarding simplethinkers signature.

I have derived the volume of a unit cube in spherical coordinates.

\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \;\ \int_{0}^{tan^{-1}(csc({\theta})} \;\ \int_{0}^{sec({\phi})}{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}


I don't believe this. The range for phi looks wrong (going only up to arctan(sqrt(2)) maximum....). I believe you will need to split into more than 1 triple integrals if you want nice limits, e.g. doing rho last you will have basically 3 parts:- one for 0<rho<1 (an eighth of sphere), one for 1<rho<sqrt(2) (spherical non-geodesic hexagon, with 3 sides geodesic), one for sqrt(2)<rho<sqrt(3) (spherical non-geodesic triangles). Those spherical integrals can be done with Gauss-Bonnet.


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PostPosted: Thu, 1 Jan 2009 19:40:22 UTC 
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cody wrote:
Hey simplethinker. I just lumbered through it from scratch. Similar to the method in

the first problem. May I ask you to demonstrate the Cauchy method you mentioned. I

was unfamiliar with it until you mentioned it.


The Cauchy Integral formula is a complex analysis means of doing things like integration.

It states that

$f^{(n)}(z)={n!\over 2\pi i}\int_\gamma{f(\xi)\over (\xi-z)^{n+1}}\,d\xi

Where \gamma is any closed, simple, rectifiable curve in the complex numbers which contains the point z.

For more information, I recommend you to the wikipedia article on the matter.

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