4

First of all I have a small request. I was banned for a very long time because I got lazy and formulated my questions without any effort. I accept the punishment and I'm happy I got a second chance. From now on I try to put as much effort in my questions as possible. This is my first question after the ban and it would make me happy if you don't down-vote if you think that my question is not good or not understandable since I'm probably getting banned again. Please write suggestions for improvement in the comments. Thank you and sorry for this way too long introduction. So here we go :)

What do I want to solve?

I want to calculate $$I = \cfrac{1}{2\pi i}\int \limits_{c-i\infty}^{c+i\infty}\cfrac{\log(s-1)}{s^2}\,x^s \mathrm{d}s,$$ where $c>1$ along the same contour as in "Applications of integral theorems", example 5: https://en.wikipedia.org/wiki/Contour_integration#Example_5_–_the_square_of_the_logarithm. To implement our integral $I$ we modify our desired closed contour integral to $$\oint_\mathcal{C}= \left(\int_{R_1}+\int_{R_2}+\int_M+\int_N+\int_r+\int\limits_{c-i\infty}^{c+i\infty}\right)\cfrac{\log(s-1)}{s^2}\,x^s\mathrm{d}s,$$ where$R_1$ ends at $c+i\infty$ and $R_2$ ends at $c-i\infty$. The whole integral is actually an Inverse laplace transform, if we define $t=\log(x)$. If we put $\frac{\log(s-1)}{s^2}$in this online calculator https://www.emathhelp.net/calculators/differential-equations/inverse-laplace-transform-calculator/?f=ln%28s-1%29%2Fs%5E2 we get $-t\operatorname{Ei}(t)+e^t-1 \Longleftrightarrow-\log x\operatorname{li}(x)+x-1$.

How I tried to solve it

The integrals $R_1,R_2$ and $r$ tend to zero (I use the same notation as in the Wikipedia link). The closed contour is also equal to zero since $0$ is excluded from the contour. After parameterizing $s=-s_\mathcal{R}+i\varepsilon$ for $M$ and $s=-s_\mathcal{R}-i\varepsilon$ for $N$ we get: $$I + \int\limits_{0}^{\infty}\cfrac{\log\lvert-s_\mathcal{R}+i\varepsilon-1\rvert + i\pi}{(-s_\mathcal{R}+i\varepsilon)^2}\,x^{-s_\mathcal{R}+i\varepsilon}\mathrm{d}s_\mathcal{R}- \int\limits_{0}^{\infty}\cfrac{\log\lvert-s_\mathcal{R}+i\varepsilon-1\rvert - i\pi}{(-s_\mathcal{R}-i\varepsilon)^2}\,x^{-s_\mathcal{R}-i\varepsilon}\mathrm{d}s_\mathcal{R} = 0$$ If we now let $\varepsilon\rightarrow0$ we can also evaluate the absolut value. Our new expression is $$\begin{align} I + \int\limits_{0}^{\infty}\cfrac{\log(s_\mathcal{R}+1) + i\pi}{s_\mathcal{R}^2}\,x^{-s_\mathcal{R}}\mathrm{d}s_\mathcal{R}- \int\limits_{0}^{\infty}\cfrac{\log(s_\mathcal{R}+1) - i\pi}{s_\mathcal{R}^2}\,x^{-s_\mathcal{R}}\mathrm{d}s_\mathcal{R} =0, \end{align}$$ which is the same as saying that $$I = -2\pi i\int\limits_{0}^{\infty}\cfrac{1}{x^{s_\mathcal{R}}s_\mathcal{R}^2}\mathrm{d}s_\mathcal{R}$$ And now the disappointment: If I put "integral from 0 to infinity of 1/(a^x*x^2)" into Wolfram Alpha it returns "(integral does not converge)". I can't find the mistake I did while evaluating. Please give me a small hint what I should correct to get the desired result. Thank you :)!

Edit: I want to thank Maxim for helping me with my question. My mistake was the definition of the branch of $\log(s-1)$. Now I (only) have to solve $$\lim_{\varepsilon\to0}\left(\int\limits_{0}^{\infty}\cfrac{\log\lvert-s_\mathcal{R}+i\varepsilon-1\rvert + i\pi}{(-s_\mathcal{R}+i\varepsilon)^2}\,x^{-s_\mathcal{R}+i\varepsilon}\mathrm{d}s_\mathcal{R}- \int\limits_{0}^{\infty}\cfrac{\log\lvert-s_\mathcal{R}+i\varepsilon-1\rvert - i\pi}{(-s_\mathcal{R}-i\varepsilon)^2}\,x^{-s_\mathcal{R}-i\varepsilon}\mathrm{d}s_\mathcal{R}\right),$$ what could get a little bit hard since I'm not allowed to set $\varepsilon$ to $0$ immediatly.

Daniel
  • 432
  • You want to show that $I = 0$ for $0 < x \leq 1$ and $I = -\ln x \operatorname {li} x + x - 1$ for $x > 1$. In the latter case, the integral over $\operatorname {Re} s = c$ is equal to the integral over a loop enclosing the ray $(-\infty, 1]$, but the subsequent steps are not justified. A simpler way to derive the transform is to start with proving that $$\int_0^\infty \operatorname {Ei}(t) e^{-s t} dt = -\operatorname {v. ! p.} \int_{-1}^\infty \frac {du} {u (u + s)}, \quad \operatorname {Re} s > 1.$$ – Maxim Oct 19 '20 at 13:10
  • Thanks for your comment! I think I understand HOW to solve the integral. But could you please tell me in a few words WHY the ray is $(-\infty,1]$ and not $(-\infty,0]$? – Daniel Oct 19 '20 at 16:38
  • 1
    Consider why there is the condition $c > 1$ (find the location of the branch cut of $\ln(s - 1)$, assuming that $\ln$ is the principal branch of the logarithm). Also consider why the integral over a semicircle of radius $r$ around $0$ does not tend to zero when $r \to 0$. – Maxim Oct 19 '20 at 18:04
  • Thanks again, I think I understand it. I'll try to solve it again. If I still have one or two questions, can I direct message you? Otherwise we have a dialogue in the comment section. – Daniel Oct 19 '20 at 18:17

1 Answers1

-1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The integral $\ds{\underline{vanishes\ out}}$ when $\ds{\underline{\Re\pars{\ln\pars{x} < 0}}}$.

When $\ds{\underline{\Re\pars{\ln\pars{x} > 0}},\quad}$ lets $\ds{\tau \equiv \ln\pars{x} >0 }$ \begin{align} I & \equiv \bbox[5px,#ffd]{\left.% \int_{c - \ic\infty}^{c + \ic\infty}\ {\ln\pars{s - 1} \over s^{2}}\expo{\tau s} {\dd s \over 2\pi\ic}\,\right\vert_{\substack{c\ >\ 1 \\[2mm] \Re\pars{\tau\ \equiv\ \ln\pars{x}}\ >\ 0}}} \\[5mm] & = -\int_{-\infty}^{1}\ {\ln\pars{1 - s} + \ic\pi \over \pars{s + \ic 0^{+}}^{2}}\expo{\tau s}{\dd s \over 2\pi\ic} \\[2mm] & \,\,\,\,- \int_{1}^{-\infty}\ {\ln\pars{1 - s} - \ic\pi \over \pars{s - \ic 0^{+}}^{2}}\expo{\tau s}{\dd s \over 2\pi\ic} \\[5mm] & = -\int_{-1}^{\infty}\ {\ln\pars{1 + s} + \ic\pi \over s^{2} - \ic\on{sgn}\pars{s}\,0^{+}} \expo{-\tau s}{\dd s \over 2\pi\ic} \\[2mm] &\,\,\,\,\,\,\, - \int_{-1}^{\infty}\ {\ln\pars{1 + s} - \ic\pi \over s^{2} + \ic\on{sgn}\pars{s}\,0^{+}} \expo{-\tau s}{\dd s \over 2\pi\pars{-\ic}} \\[5mm] & = -2\,\Re\int_{-1}^{\infty}\ {\ln\pars{1 + s} + \ic\pi \over s^{2} - \ic\on{sgn}\pars{s}\,0^{+}} \expo{-\tau s}{\dd s \over 2\pi\ic} \\[5mm] & = -{1 \over \pi}\,\Im\int_{-1}^{\infty}\ {\ln\pars{1 + s} - \ic\pi \over s^{2} - \ic\on{sgn}\pars{s}\,0^{+}} \expo{-\tau s}\dd s \\[5mm] & = -{1 \over \pi}\,\Im\left\{% \mrm{P.V.}\int_{-1}^{\infty}\ {\ln\pars{1 + s} - \ic\pi \over s^{2}} \expo{-\tau s}\dd s \right. \\[2mm] &\ \left.% + \int_{-1}^{\infty} \bracks{\ln\pars{1 + s} - \ic\pi}\expo{-\tau s} \bracks{\ic\pi\on{sgn}\pars{s}\,\delta\pars{s^{2}}}\,\dd s\right\} \\[5mm] & = \mrm{P.V.}\int_{-1}^{\infty}{\expo{-\tau s} \over s^{2}}\,\dd s - {1 \over 2} \int_{-1}^{\infty}{\ln\pars{1 + s} \over s} \expo{-\tau s}\,\delta\pars{s}\,\dd s \\[5mm] & = 2\ \underbrace{\int_{0}^{1} {\cosh\pars{\tau s} \over s^{2}}\,\dd s} _{\ds{\color{red}{\mbox{It diverges !!!}}}} \\[2mm] & +\ \underbrace{\int_{1}^{\infty}{\expo{-\tau s} \over s^{2}}\,\dd s} _{\ds{{1 \over x} - \ln\pars{x}\on{E}_{1}\pars{\ln\pars{x}}}}\ -\ {1 \over 2} \end{align}

$\ds{\on{E}_{1}}$ is the Exponential Integral Function.

Felix Marin
  • 89,464
  • 1
    The integrand doesn't have any singularities on a loop around $(-\infty, 1]$ and decays exponentially when $\operatorname {Re}s \to -\infty$, clearly the integral over the loop converges. – Maxim Oct 20 '20 at 21:48
  • Are you saying that the integral either vanishes or diverges? Please read my question again - I already know that the integral converges but I didn't know how to evaluate it correctly. – Daniel Oct 20 '20 at 22:05
  • @Daniel How do you know that the integral converges ?. Do you have a proof ?. – Felix Marin Oct 21 '20 at 18:40
  • @Maxim The problem isn't with the loop: It's with the behavior at $\displaystyle s = 0$. – Felix Marin Oct 21 '20 at 18:42
  • @Felix Marin First of all with Maxim I was able to proof that it converges and secondly we don't care about the case $s=0$ because it's not in or on our contour. We even excluded it on purpose. – Daniel Oct 21 '20 at 18:58
  • By loop, I mean the whole contour, say $(-\infty - i, 2 - i, 2 + i, -\infty + i)$. You've obtained an integral over $(-\infty, -\epsilon] \cup [\epsilon, 1]$ and considered the limit for $\epsilon \to 0$ (the p.v.), which is infinite. The problem with your derivation is indeed that the limit of the integral over the parts of the contour connecting $-\epsilon$ to $\epsilon$ is also infinite. – Maxim Oct 21 '20 at 19:57
  • @Maxim I don't have "contour connecting $-\epsilon$ to $\epsilon$". There is a "piece" which I omitted because it vanishes out: It's the indent around $1$: $$ -\int_{\pi}^{-\pi}{\ln\left(\epsilon\right) + \mathrm{i}\theta \over 1^{2}}\epsilon,\mathrm{e}^{\mathrm{i}\theta},,\mathrm{i},{\mathrm{d}\theta \over 2\pi\mathrm{i}} $$ which vanishes out as $\epsilon \to 0^{+}$. – Felix Marin Oct 28 '20 at 02:41
  • You start with an integral $I$ over a continuous curve enclosing the ray $(-\infty, 1]$ and then consider the limit of an integral $I_\epsilon$ over $(-\infty, -\epsilon] \cup [\epsilon, 1]$. The latter contour is not a continuous curve. If you work out how you go from $I$ to $I_\epsilon$, you'll find out that the difference between them doesn't have a finite limit. This may help: consider $x = 1$. Then $I = 0$ because Cauchy's integral theorem applies. Your derivation still gives infinities. – Maxim Oct 28 '20 at 14:14