The Bohr-Mollerup Theorem [Fixed tech difficulties, thx]

In summary, the conversation is about a mathematical problem that the person is trying to solve and asking for help in finding their error or getting guidance on how to finish it correctly. They share their work so far and mention that they believe there may be an issue with their second application of integration by parts.
  • #1
benorin
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Homework Statement
Ok, so for May 2020's Math Challenge (last month), problem #3, fresh_42 made room for an alternative solution (a completely real variable proof) for the integral ##\int_{-\infty}^{\infty}\tfrac{\cos (\alpha x)}{1+x^2}\, dx= \pi e^{-\alpha}\quad (\alpha \geq 0)##.
Relevant Equations
I've since learned how to solve this problem by the differiating under the integral sign technique, and that is a solution to be sure, but I've got an idea to use the Bohr-Mollerup Theorem here and I realize it's not the most practical way to do the problem, but I'd like to see if *we* can get it to work.This is the theorem statement:

The Bohr-Mollerup Theorem (1) any function that has the functional equation ##f(x+1)=xf(x)## over the real interval ##(0,\infty )##, (2) the value ##f(1)=1##, and (3) is log-convex is the Gamma function ##\Gamma (x)##.
Now I realize this is not the simplest way to do this problem, I get that, so please don't answer me with the "Try doing it this way..." posts. I would like to see if we can please make this solution come to life. The first kink in the proof is the functional equations, I know it should work, because I know the value of the integral already, but integrals can have convergence issues to be sure. When I came up with this work I felt as if I had ignored something my analysis prof taught me, somewhere, right? You're good at this. please help me discover my error(s) or else show me how to finish this off correctly because I have some infinities coming up where there should be a zero. Here's what I have so far, mind you that the functional equation part is merely the first challenge, and that I expect the log-convexity may prove difficult too.

Off the bat using symmetry to fix the domain of interest, define $$I(\alpha ):=2\int_{0}^{\infty}\tfrac{\cos (\alpha x)}{1+x^2}\, dx$$ and $$f(\beta ):= \tfrac{1}{\pi }\int_{0}^{\infty} t^{\beta -1} I(t)\, dt =\tfrac{2}{\pi }\int_{0}^{\infty}\int_{0}^{\infty} t^{\beta -1} \tfrac{\cos (t x)}{1+x^2}\, dt dx$$

where ##f(\beta )## is our candidate for ##\Gamma (\beta )## and the double integral (being always positive) is abs. convergent so Fubini's Theorem applies, going for the simplest route to functional equation I figure to keep it nice for us I will instead prove that ##f(\beta +2)=(\beta +1)\beta f(\beta )## (which is just as good, pretty sure) by doing integration by parts twice. Here's the work:

$$f(\beta +2) = \tfrac{2}{\pi }\int_{0}^{\infty}\int_{0}^{\infty} t^{\beta +1} \tfrac{\cos (t x)}{1+x^2}\, dt dx$$

scratch_1: ##u_1=t^{\beta +1}\implies du_1=(\beta +1)t^{\beta} dt\text{ and } dv_1 = \tfrac{\cos (t x)}{1+x^2}dt\implies v_1=\tfrac{\sin (t x)}{x(1+x^2)}##

and the integral becomes

$$f(\beta +2) = \tfrac{2}{\pi }\int_{0}^{\infty}\left[ t^{\beta +1} \tfrac{\sin (t x)}{x(1+x^2)}\right| _{t=0}^{\infty }\, dx - \tfrac{2}{\pi }(\beta +1) \int_{0}^{\infty}\int_{0}^{\infty} t^{\beta} \tfrac{\sin (t x)}{x(1+x^2)}\, dt dx$$

scratch_2: ##u_2=(\beta +1) t^{\beta}\implies du_2=(\beta +1)\beta t^{\beta -1} dt\text{ and } dv_2 = \tfrac{\sin (t x)}{x(1+x^2)}dt\implies v_2=-\tfrac{\cos (t x)}{x^2 (1+x^2)}##

and we get (evaluating the ##t##'s since the ##t=0## gives a zero term for both) and partial fraction decompositions next,

$$\begin{gathered} f(\beta +2) = \tfrac{2}{\pi }\lim_{t\to\infty}\left\{ t^{\beta +1}\int_{0}^{\infty}\sin (t x)\left( \tfrac{1}{x}-\tfrac{x}{1+x^2}\right) \, dx - (\beta +1)\beta t^{\beta}\int_{0}^{\infty}\cos (t x)\left( \tfrac{1}{x^2}-\tfrac{1}{1+x^2}\right) \, dx\right\} \\ - \tfrac{2}{\pi} (\beta +1)\beta \int_{0}^{\infty} \int_{0}^{\infty} t^{\beta -1} \cos (t x)\left( \tfrac{1}{x^2}-\tfrac{1}{1+x^2}\right) \, dt dx\\ \end{gathered}$$

separating terms, and using the substitution ##u=tx\implies du = t dx## in the first term of the partial fractions the resulting Sine Integral and Cosine Integral-like functions and using the derivative obtained by differentiation under the integral sign of ##I (t)##, namely ##\tfrac{\partial I}{\partial t} =-\tfrac{2}{\pi}\int_{0}^{\infty} \tfrac{x\sin (tx)}{1+x^2}\, dx## and the first dbl integral term is separable under the same change of variables ##u=tx## so we have

$$ \begin{gathered} f(\beta +2)= \tfrac{2}{\pi }\lim_{t\to\infty}\left\{ t^{\beta +3}\underbrace{ \text{Si} (t)}_{\to\tfrac{\pi}{2}\text{ as } t\to\infty} +\underbrace{t^{\beta +1}\tfrac{\partial I}{\partial t}}_{\to 0\text{ as } t\to\infty} + (\beta +1) t^{\beta +3}\int_{0}^{\infty}\tfrac{\cos (u)}{u^2} \, du-t^{\beta} I(t) \right\} \\ - \tfrac{2}{\pi} (\beta +1)\beta \int_{0}^{\infty}t^{\beta +2}\, dt \int_{0}^{\infty} \tfrac{ \cos ( u)}{u^2} du + (\beta +1)\beta f(\beta ) \\ = \tfrac{2}{\pi }\lim_{t\to\infty}\left\{ t^{\beta +3}\tfrac{\pi}{2} + 3\left( \tfrac{\beta +1}{\beta +3}\right) t^{\beta +3}\int_{0}^{\infty}\tfrac{\cos (u)}{u^2} \, du-t^{\beta} I(t) \right\} \\ - \tfrac{2}{\pi} (\beta +1)\beta \int_{0}^{\infty}t^{\beta +2}\, dt \int_{0}^{\infty} \tfrac{\cos ( u)}{u^2} du + (\beta +1)\beta f(\beta ) \\ \end{gathered} $$

where ##\text{Si} (t)## is the Sine Integral function. The very last term should be the final result, ## (\beta +1)\beta f(\beta )##, and all else should vanish, but when last I worked this problem I remember thinking this was infinity and not zero. Thanks for your time,

-Ben

Note: I'm having cache issues and I cannot preview my TeX w/o submitting this post. I will submit this now because last I knew I didn't have any typesetting errors, or at least everything displayed, I'm not positive all the math is correct though.
 
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  • #2
I think I found the error, the second application of int by parts is invalid bc it's not abs convergent
 

Related to The Bohr-Mollerup Theorem [Fixed tech difficulties, thx]

1. What is the Bohr-Mollerup Theorem?

The Bohr-Mollerup Theorem is a mathematical theorem that states that the gamma function is the only function that satisfies three key properties: it is continuous, logarithmically convex, and takes on the value 1 at the number 1. It was first proven by physicists Niels Bohr and Johannes Mollerup in the early 20th century.

2. What is the significance of the Bohr-Mollerup Theorem?

The Bohr-Mollerup Theorem has significant implications in mathematics and physics. It helps to establish the uniqueness of the gamma function and its importance in various areas of mathematics, such as complex analysis and number theory. It also has applications in physics, particularly in quantum mechanics and statistical mechanics.

3. How does the Bohr-Mollerup Theorem relate to the gamma function?

The Bohr-Mollerup Theorem states that the gamma function is the only function that satisfies three specific properties, making it a unique function. This is important because the gamma function is widely used in mathematics and physics, and the theorem helps to establish its significance and properties.

4. Are there any real-world applications of the Bohr-Mollerup Theorem?

Yes, the Bohr-Mollerup Theorem has various real-world applications in mathematics and physics. For example, it is used in probability theory, where the gamma function is used to model continuous probability distributions. It also has applications in statistical mechanics, where it is used to calculate the partition function of a system.

5. Is there any controversy surrounding the Bohr-Mollerup Theorem?

No, the Bohr-Mollerup Theorem is a well-established and widely accepted theorem in mathematics and physics. It has been rigorously proven and has numerous applications, making it an important and uncontroversial result in the scientific community.

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