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I am trying to prove that $$ \frac{1}{2\pi i}\int^{a+i\infty}_{a-i\infty}\frac{x^s}{s-\beta}ds =\begin{cases} x^{\beta}, & x > 1 \\ 0, & 0 < x < 1 \\ \end{cases} $$ for $0<{\rm Re}(\beta)<a$ by using Feynman integration and solving a homogeneous second order differential equation respectively. Here is my attempt: \begin{align*} \frac{1}{2\pi i}\int^{a+i\infty}_{a-i\infty}\frac{x^s}{s}ds& =\frac{x^a}{2\pi}\int^{\infty}_{0}\frac{x^{it}(a-it)+x^{-it}(a+it)}{a^2+t^2}dt\\ & =\frac{x^a}{\pi}\left(\int^{\infty}_{0}\frac{\cos(\ln(x)ta)}{1+t^2}dt+\int^{\infty}_{0}\frac{t\sin(\ln(x)ta)}{1+t^2}dt\right). \end{align*} \begin{align*} {\rm Re}(\oint_C\frac{x^{iza}}{1+z^2}dz) & ={\rm Re}\left(\lim_{R \to \infty}\int^{R}_{-R}\frac{x^{iza}}{1+z^2}dz+\int_{\Gamma}\right) \\ & =\frac{\pi}{x^a}.\lim_{R \to \infty} \left| \int_{\Gamma} \right| \\ & \leqslant\lim_{R \to \infty} \frac{2}{R}\int^{\frac{\pi}{2}}_{0}\frac{x^{-2R\theta/\pi}}{1+(Re^{i\theta})^{-2}}d\theta \\ & =0. \end{align*} Therefore $$ \int^{\infty}_{0}\frac{\cos(\ln(x)ta)}{1+t^2}dt = \frac{\pi}{2x^a}. $$ Set $$ I(a):=\int_{R}\frac{\cos(\ln(x)ta)}{1+t^2}dt $$ so $$ I'(a)=\ln(x)(-{\rm sgn}(\ln(x))\pi+\int_{R}\frac{\sin(\ln(x)ta)}{t(1+t^2)}dt) $$ and $$ I''(a)-\ln^2(x)I(a) = 0. $$ Therefore $I(a)=c_1x^a+c_2x^{-a}$ for $$ c_1=\begin{cases} 0, & x > 1 \\ \pi, & 0 < x < 1 \\ \end{cases} \text{ and } c_2= \begin{cases} \pi, & x > 1 \\ 0, & 0 < x < 1.\end{cases} $$

Assuming my attempt is correct thus far, can I evaluate $$ \int_{R}\frac{t\sin(\ln(x)at)}{1+t^2}dt=-\frac{I'(a)}{\ln(x)}:=D(a)={\rm sgn}(\ln(x))\pi-\int_{R}\frac{\sin(\ln(x)at)}{t(1+t^2)}dt. $$ Then $D'(a)=-\ln(x)I(a)$, so $$D(a)=c_2x^{-a}-c_1x^{a}.$$ $$ \frac{1}{2\pi i}\int^{a+i\infty}_{a-i\infty}\frac{x^s}{s}ds=\frac{x^{a}}{2\pi}((c_1x^{a}+c_2x^{-a})+c_2x^{-a}-c_1x^{a})=\begin{cases} 1, & x > 1 \\ 0, & 0 < x < 1. \end{cases} $$ (as defined herein) using Feynman integration? It is used extensively in Riemann's paper in finding an analytic representation of the jump function (equivalently, an asymptotic estimate of the prime counting function by applying the Moebius inversion formula [following from the associativity of the Dirichlet convolution] to the analytic representation of the jump function) and in a proof of Perron's formula. Pardon my illegible uglyography and quotidian rogitation, albeit I consider the transient apanthropinization as a pars pro toto for expunged pedagogical anomalies.

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    $\begingroup$ Is this a serious post? The "precalculus" tag makes absolutely no sense. Anyway, your original integral can be simplified by dividing by $x^\beta$ and making the change of variables $s \mapsto s+\beta$ and rewriting $a-\beta$ as $a$, so you want to show $(1/(2\pi i))\int_{a-i\infty}^{a+i\infty} (x^s/s)\,ds$ for ${\rm Re}(a) > 0$ is $1$ if $x > 1$ and $0$ if $0 < x < 1$. So your $\beta$ is a fake parameter in the calculation. I don't know if this can be computed using differentiation under the integral sign, but have you read a proof of the integral anywhere (an analytic number theory book)? $\endgroup$
    – KCd
    Jul 26, 2021 at 4:17
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    $\begingroup$ Thanks for putting more meaningful tags on the problem. Your typesetting sucked in some places (you admitted that yourself :)) so I made edits to make it easier to read. (Don't abbreviate words like "Therefore" to $\therefore$ when typing math: nobody does such things in printed material.) The logical connections from one step to the next is missing in some steps. Please reread your post and insert connecting words to explain how to go from each step to the next. For instance, the real part of a contour integral comes out of nowhere early on. $\endgroup$
    – KCd
    Jul 26, 2021 at 17:16
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    $\begingroup$ Your desired calculation $\int_0^\infty (\cos((\ln x)ta)/(1+t^2))\,dt \stackrel{?}{=} \pi/(2x^a)$ is correct if $(\ln x)a > 0$ (not if $(\ln x)a < 0$), which is equivalent to requiring $x^a > 1$. With the change of variables $b = (\ln x)a > 0$, the equation can be rewritten as $\int_0^\infty (\cos(bt)/(1+t^2))\,dt \stackrel{?}{=} (\pi e^{-b})/2$ for $b > 0$. This is proved using differentiation under the integral sign using a second-order ODE in kconrad.math.uconn.edu/blurbs/analysis/diffunderint.pdf. See Section 14, especially equation (14.1). Such an approach is not simple. $\endgroup$
    – KCd
    Jul 26, 2021 at 17:25
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    $\begingroup$ Don't use $\wedge$ for "and" when typing math. Unless you are writing a paper on mathematical logic, use words in place of logical symbols such as $\wedge$ or $\forall$ or $\exists$. In math textbooks and papers you'll see people basically rarely write such symbols unless the topic is about logic (unlike blackboard writing). $\endgroup$
    – KCd
    Jul 26, 2021 at 17:29
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    $\begingroup$ Interesting final sentence $\endgroup$
    – FShrike
    Jul 23, 2023 at 0:01

1 Answer 1

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Substituting $s=a+it$ in your integral gives $$ \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{it \log x}}{t-i(a-\beta)} \, dt = \begin{cases} x^{\beta-a}, \; x>1 \\ 0, \; 0<x<1 \end{cases} $$ and under a change of variables we see it is equivalent to the formula $$ \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{ixt}}{t-w} \, dt = \begin{cases} e^{iwx}, \; x>0 \\ 0, \; x<0 \end{cases} \hspace{1cm} (*) $$ where $x$ is real and non-zero and $w \in \mathbb{H}$. To apply the Feynman technique define the function $$ F(x) = \int_0^{\infty} \frac{\sin(xt)}{t(b^2+t^2)} \, dt, \hspace{0.5cm} x>0 $$ where $b>0$ is real. The intuition is we need a function that decreases sufficiently fast that two derivatives under the integral sign may be taken (check this) thus leading to a second order differential equation. Differentiating twice gives $$ F''(x) = \int_0^{\infty} \frac{-t^2\sin(xt)}{t(b^2+t^2)} \, dt = \int_0^{\infty} \left(\frac{b^2}{t(b^2+t^2)} - \frac{1}{t} \right) \sin(xt) \, dt = b^2 F(x) - \frac{\pi}{2} $$ and solving this differential equation gives $$ F(x) = C_1 e^{bx} + C_2 e^{-bx} + \frac{\pi}{2b^2}, \hspace{0.5cm} x>0 $$ As $x \to \infty, F(x)$ remains bounded so $C_1=0$, and as $x \to 0^+, F(x) \to 0$ so we require $C_2=-\frac{\pi}{2b^2} $. Therefore $F(x) = \frac{\pi}{2b^2}-\frac{\pi}{2b^2} e^{-bx}$ for $x>0$, and differentiating under the integral once and twice gives the pair $$ \int_0^{\infty} \frac{\cos(xt)}{b^2+t^2} \, dt = \frac{\pi}{2b} e^{-bx}, \hspace{1cm} \int_0^{\infty} \frac{t\sin(xt)}{b^2+t^2} \, dt = \frac{\pi}{2} e^{-bx} $$ for $x>0$. Of course the first integral is an even function of $x$ and the second is an odd function of $x$ so we must in fact have $$ \int_0^{\infty} \frac{\cos(xt)}{b^2+t^2} \, dt = \frac{\pi}{2b} e^{-b|x|}, \hspace{1cm} \int_0^{\infty} \frac{t\sin(xt)}{b^2+t^2} \, dt = \frac{\pi}{2} \text{sgn}(x) e^{-b|x|}, \hspace{0.5cm} x \in \mathbb{R} \setminus \{0\} $$ We now combine these formulas to obtain $(*)$. Letting $w=a+ib$ for $a \in \mathbb{R}$ and $b>0$ write $$ \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{ixt}}{t-(a+ib)} \, dt = \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{ix(t+a)}}{t-ib} \, dt = \frac{e^{iax}}{2\pi i} \int_{-\infty}^{\infty} \frac{(t+ib) e^{ixt}}{t^2+b^2} \, dt $$ $$ = \frac{e^{iax}}{2\pi i} \left(2i \int_0^{\infty} \frac{t\sin(xt)}{b^2+t^2} \, dt + 2ib \int_0^{\infty} \frac{\cos(xt)}{b^2+t^2} \, dt \right) $$ $$ = \frac{e^{ia}}{2} \left(\text{sgn}(x) e^{-b|x|} + e^{-b|x|} \right) = \begin{cases} e^{i(a+ib)x}, \; x>0 \\ 0, \; x<0 \end{cases} $$

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