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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

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

*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.