Fun With Integrals

[Nootropic]

So I was bored in one of my classes (probably physics), so I started looking at some patterns in integrals. I was looking at a way of integrating ((1-e^(x))^n by means of a rationalizing substitution (because who wants to use the binomial theorem?), and then I continued on through a various list of random functions I had in mind, and came to the logarithm function log(x), or ln(x), if you prefer that notation.

 

Now instead of 1 - e^(x), I inserted 1 - log(x), to obtain ((1-log(x))^n, and now for the substitution: 1 - log(x) = u , so e^(1-u) = x and dx = -e^(1-u) du, inserting these into the integral I forgot to mention earlier, we obtain the integral of -u^n *e^(1-u). Now this looks a lot like the gamma function (actually it is -e *gamma(n+1)). And I also noticed that (1-log(x))^n would create a series representation of for the function 1/log(x) (albeit convergent but not equal to the function 1/log(x), but maybe a decent asymptotic expansion). This is simply done by inserting 1 - log(x) into the function 1/1-x. So basically what I'm trying to get at is there some kind of connection between the gamma function and the logarithmic integral? It remains to be revealed...

 

Logarithmic integral - Wikipedia, the free encyclopedia

 

Gamma function - Wikipedia, the free encyclopedia

[LaurieAG]

So basically what I'm trying to get at is there some kind of connection between the gamma function and the logarithmic integral? It remains to be revealed...

 

Logarithmic integral - Wikipedia, the free encyclopedia

 

Gamma function - Wikipedia, the free encyclopedia

 

Hello Nootropic,

 

For a start the plots of the Logarithmic integral function Logarithmic integral function - Wikipedia, the free encyclopedia is used on prime numbers and is totally different to the Gamma function plot which is an extension of the factorial function to real and complex numbers.

[Nootropic]

oh, I know. I'm well aware of the prime number theorem and what the gamma function is. I'm just wondering if there is some kind of connection, as I may be observing. One finds all kinds of strange connections in mathematics, as I'm sure you know.

[Nootropic]

I guess it is of interest to note that that (1 - ln(x))^n is easy to integrate using parts, though making the substitution u = 1- ln(x) leads to an intersting result.

[LaurieAG]

oh, I know. I'm well aware of the prime number theorem and what the gamma function is. I'm just wondering if there is some kind of connection, as I may be observing. One finds all kinds of strange connections in mathematics, as I'm sure you know.

 

Hi Nootropic,

 

What you have come across is probably another example of paralogic/paralogism.

 

Par`a*log"ic*al, a. Containing paralogism; illogical. ``Paralogical doubt.'' --

 

In the past couple of years there was a good example of this with regards to using atomic elements of massively unequal weights to gain a plot that was similar (but shouldn't have been due to the weight discrepancy). They called it 'Para?????' . Maybe you have stumbled onto something similar?

[Nootropic]

Actually, not quite. Let's look at the gamma function. It is defined as the integral of e^(-t)t^(x-1) dt with limits x = 0 and x = + infinity. Let's make a substitution: e^(-t) = u, then -ln(u) = t and -1/t dt = du, so changing the integral around we end up with the integral of -(ln(u))^(x-1). Now the limits can be transformed by taking the value of of e^(-t) at them. The limit as t approaches infinity is simply zero and the value at zero is one. And since this is the opposite of the integral we can rewrite the integral as the integral of (ln(u))^(x-1) from x = 0 to x = 1. Viola! And I knew I wasn't crazy, I just can't remember where I had seen this before, but I actually thought of all this until halfway through this post, I remembered it was in my book about the Riemann Zeta Function. Euler wrote out the gamma function in terms of the logarithm, but integrating with respect to y, so he was integrating the exponential function.

 

Paralogism, I think not!

[Nootropic]

Though, of course, there is still the question whether or not the function 1/ln(x) has a valid expansion of any kind. It's easily seen that f(0) = 0, and I haven't taken the liberty to really look at the first and second derivatives, though my guess would be that it would have a valid mclaurin expansion, but not equal the function, such as the function e^(-1/x^2). And then the question would be how it relates to the gamma function.

[Qfwfq]

First, you should really say the limit rather than the value in zero: [math]\lim_{x\rightarrow0}\frac{1}{\ln x}=0[/math] and then second, the derivative isn't so hard. [math]\frac{d}{dx}\frac{1}{\ln x}=-\frac{1}{x\ln^2 x}[/math] and by de l'Hopital I get [math]\lim_{x\rightarrow0}\frac{1}{x\ln^2 x}=\lim_{x\rightarrow0}\frac{1}{2x}[/math] and third it would be easier for folks to read you if you cared to use LaTeX.

[Nootropic]

oh, yeah the limit thing is true. I know what the derivative is, and is easy to calculate, I just hadn't bothered with l'hopital's rule because it wasn't something I was actively working on at the moment. It would be user to write a program to calculate the nth derivative, obviously. Maybe I'll figure out a way to do that.

[Nootropic]

Your simplification of the limit using l'hopital's rule is slightly incorrect. It should be lim x --> - 1/2x. Also, look at a graph, you can clearly see it approaches negative infinity as x approaches zero.

[Qfwfq]

Actually, I had paid attention to the signs. Perhaps something escaped your notice. :)

[Nootropic]

Hm...don't know. What am I missing?

[Qfwfq]

It's obvious that the derivative is negative, even without looking at the graph. I did not say otherwise. :)

[Nootropic]

Well, yeah, it just seemed a strange way of going about things...

[LaurieAG]

Paralogism, I think not!

 

Actually Nootropic, if the plots of the functions were identical (or even remotely close) you could argue that paralogism wasn't involved.

 

BTW, the example I gave before may not be a para???? because recent research (sorry I cannot find a reference on the web, found it!) has indicated that a previously thought neutral charge may in fact be the result of a positive and negative charge cancelling each other out across each cycle.

 

http://hypography.com/forums/general-science-news/12841-research-overturns-accepted-notion-neutrons-electrical.html#post190125