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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.
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.