In THIS ANSWER, I used straightforward complex analysis to show that

$$\gamma =2\int_0^\infty \frac{\cos(x^2)-\cos(x)}{x}\,dx \tag 1$$

where $\gamma =-\int_0^\infty \log(x) e^{-x}\,dx$ is the Euler-Mascheroni Constant.

The key in the derivation of $(1)$ was to transform the cosine terms into real exponential ones.

To date, I have been unable to use strictly real analysis, without appealing to tabulated results of special functions (e.g., use of the $\text{Cin}(x)$ and $\text{Ci}(x)$ functions), to prove $(1)$.

I have tried introducing a parameter and using "Feynman's Trick to augment the integral into something manageable. Or somewhat equivalently, rewriting the integral in $(1)$ as a double integral and proceeding by exploiting Fubini-Tonelli.

QUESTION: What are ways to prove $(1)$ without relying on complex analysis and without simply appealing to tabulated relationships of special functions. For example, stating that the $Ci(x)$ function is defined as $\text{Ci}(x)\equiv -\int_x^\infty\frac{\cos(t)}{t}\,dt=\gamma +\log(x) +\int_0^x \frac{\cos(t)-1}{t}\,dt$ is unsatisfactory unless one proves the latter equality.

  • 4
    $\begingroup$ Is bold text necessary? $\endgroup$ – Simply Beautiful Art Jan 16 '17 at 19:08
  • 2
    $\begingroup$ @SimpleArt Yes. Your answer must appear in bold-faced font; the color and style are up to you. ;-)) $\endgroup$ – Mark Viola Jan 16 '17 at 19:49
  • 1
    $\begingroup$ Ah, I see. So I may do it using \color{#FFFFFF}{}? $\endgroup$ – Simply Beautiful Art Jan 16 '17 at 19:50
  • 1
    $\begingroup$ @SimpleArt That would be an excellent choice. $\endgroup$ – Mark Viola Jan 16 '17 at 19:52

It turns out that we have the following observation:

Observation. For a nice function $f : [0,\infty) \to \Bbb{C}$, we have

$$ \int_{\epsilon}^{\infty} \frac{f(x)}{x} \, dx = -f(0)\log\epsilon + c(f) + o(1) \qquad \text{as } \epsilon \to 0^+ \tag{1} $$

where the constant $c(f)$ is computed by

$$ c(f) = \lim_{R\to\infty}\left( \int_{0}^{R} \mathcal{L}f(s) \, ds - f(0)\log R\right) - f(0)\gamma. \tag{2} $$

The reasoning is surprisingly simple: First, define $g(x) = (f(x) - f(0)\mathbf{1}_{(0,1)}(x))/x$ and notice that

$$ \int_{\epsilon}^{\infty} \frac{f(x)}{x} \, dx = -f(0)\log\epsilon + \int_{\epsilon}^{\infty} g(x) \, dx. $$

Assuming that the LHS of $\text{(1)}$ exists for all $\epsilon > 0$ and that $f$ behaves nice near $x = 0$, this implies $\text{(1)}$. Next, notice that $c(f) = \mathcal{L}g(0)$ and that $-(\mathcal{L}g(s))' = \mathcal{L}f(s) - f(0) (1-e^{-s})/s$. Therefore

\begin{align*} c(f) &= \lim_{R\to\infty} \int_{0}^{R} (-\mathcal{L}g(s))' \, ds \\ &= \lim_{R\to\infty} \left( \int_{0}^{R} \mathcal{L}f(s) \, ds - f(0) (1 - e^{-R})\log R + f(0) \int_{0}^{R} e^{-s}\log s \, ds \right) \\ &= \lim_{R\to\infty} \left( \int_{0}^{R} \mathcal{L}f(s) \, ds - f(0) \log R \right) - f(0)\gamma. \end{align*}

At this moment this is just a heuristic computation. For a broad class of functions for which the LHS of $\text{(1)}$ exists, however, this computation can be made rigorous. This is particularly true for our function $f(x) = \cos x$. Now plugging $\mathcal{L}f(s) = \frac{s}{s^2+1}$ shows that $c(f) = -\gamma$ and thus

$$ \int_{\epsilon}^{\infty} \frac{\cos x}{x} \, dx = -\log\epsilon - \gamma + o(1). $$

Plugging this asymptotics, we have

$$ \int_{\epsilon}^{\infty} \frac{\cos(x^2) - \cos x}{x} \, dx = \frac{1}{2}\int_{\epsilon^2}^{\infty} \frac{\cos x}{x} \, dx - \int_{\epsilon}^{\infty} \frac{\cos x}{x} \, dx = \frac{1}{2}\gamma + o(1) $$

and the identity follows by letting $\epsilon \to 0^+$.

Here, the constant $c(f)$ can be thought as a regularized value of the divergent integral $\int_{0}^{\infty} \frac{f(x)}{x} \, dx$. This has the following nice properties (whenever they exist)

  • $c$ is linear: $c(\alpha f(x) + \beta g(x)) = \alpha c(f) + \beta c(g)$.
  • $c(f(x^p)) = \frac{1}{p}c(f)$ for $p > 0$,
  • $c(f(px)) = c(f) - f(0)\log p$ for $p > 0$,

Together with some known values, we can easily compute other types of integrals. For instance, using the fact that $c(\cos x) = -\gamma$ and $c(e^{-x}) = -\gamma$, we have

\begin{align*} \int_{0}^{\infty} \frac{\cos (x^p) - \exp(-x^q)}{x} \, dx &= c\left\{ \cos (x^p) - \exp(-x^q) \right\} \\ &= \frac{1}{p}c(\cos x) - \frac{1}{q}c(e^{-x}) = \gamma\left( \frac{1}{q} - \frac{1}{p}\right) \end{align*}

for $p, q > 0$.

  • 2
    $\begingroup$ Sangchul, you're quite welcome. And thank you again for taking time to post this. It's a very nice generalization, which as you stated, is quite simple. -Mark $\endgroup$ – Mark Viola Feb 22 '17 at 16:53

For $$\Gamma '\left ( x \right )=\int_{0}^{\infty }e^{-t}t^{x-1}\ln t\, \mathrm{d}t$$ using $$\ln t=\int_{0}^{\infty }\frac{e^{-s}-e^{-ts}}{s}\, \mathrm{d}s$$ we have $$\Gamma '\left ( x \right )=\int_{0}^{\infty }e^{-t}t^{x-1}\int_{0}^{\infty }\frac{e^{-s}-e^{-ts}}{s}\, \mathrm{d}s\mathrm{d}t=\Gamma \left ( x \right )\int_{0}^{\infty }\left ( e^{-s}-\frac{1}{\left ( s+1 \right )^{x}} \right )\frac{\mathrm{d}s}{s}$$ Hence, let $x=1$ we get $$\gamma =\int_{0}^{\infty }\left ( \frac{1}{s+1 }-e^{-s} \right )\frac{\mathrm{d}s}{s}$$ let $s=t^k,~k>0$, we get $$\gamma =\int_{0}^{\infty }\left ( \frac{1}{t^{k}+1 }-e^{-t^{k}} \right )\frac{k\, \mathrm{d}t}{t}$$ So,let $k=a,b$ $$\frac{\gamma}{a} =\int_{0}^{\infty }\left ( \frac{1}{t^{a}+1 }-e^{-t^{a}} \right )\frac{ \mathrm{d}t}{t}~~,~~\frac{\gamma}{b} =\int_{0}^{\infty }\left ( \frac{1}{t^{b}+1 }-e^{-t^{b}} \right )\frac{ \mathrm{d}t}{t}$$ hence $$\frac{\gamma}{b}-\frac{\gamma}{a} =\int_{0}^{\infty }\left [\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )+\left ( e^{-x^a}-e^{-x^b} \right ) \right ]\frac{ \mathrm{d}t}{t}$$ then $$\int_{0}^{\infty }\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}=\int_{0}^{1}\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}+\int_{1}^{\infty }\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}$$ let $t\rightarrow \dfrac{1}{t}$,we get $$\int_{0}^{1}\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}=-\int_{1}^{\infty }\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}$$ So $$\int_{0}^{\infty }\left ( \frac{1}{t^{b}+1 }- \frac{1}{t^{a}+1 } \right )\frac{\mathrm{d}t}{t}=0$$ and $$\left ( \frac{1}{b}-\frac{1}{a} \right )\gamma =\int_{0}^{\infty }\frac{e^{-t^a}-e^{-t^b}}{t}\, \mathrm{d}t\tag1$$ Lemma:

$$\int_{0}^{\infty }\frac{e^{-t^a}-\cos t^a}{t}\, \mathrm{d}t=0~,~a>0$$

Proof: Let $$f\left ( x \right )=\int_{0}^{\infty }\frac{e^{-t}-\cos t}{t}\, e^{-xt}\, \mathrm{d}t$$ so $$f'\left ( x \right )=\int_{0}^{\infty }\left ( \cos t-e^{-t} \right )e^{-xt}\, \mathrm{d}t=\frac{x}{1+x^2}-\frac{1}{1+x}$$ hence $$\int_{0}^{\infty }f'\left ( x \right ) \mathrm{d}x=\ln\frac{\sqrt{1+x^2}}{1+x}\Bigg|_{0}^{\infty }=0=f\left ( \infty \right )-f\left ( 0 \right )$$ It's easy to see that $f\left ( \infty \right )=0$,so $$f\left ( 0 \right )=\int_{0}^{\infty }\frac{e^{-t}-\cos t}{t}\, \mathrm{d}t=0$$ Let $x^a\to t$, we get $$a\int_{0}^{\infty }\frac{e^{-t^{a}}-\cos t^{a}}{t}\, \mathrm{d}t=0\Rightarrow \int_{0}^{\infty }\frac{e^{-t^{a}}-\cos t^{a}}{t}\, \mathrm{d}t=0\tag2$$ Now using $(1)$ and $(2)$, we get $$\Large\boxed{\color{Blue} {\int_{0}^{\infty }\frac{\cos x^{a}-\cos x^b}{x}\, \mathrm{d}x=\left ( \frac{1}{b}-\frac{1}{a} \right )\gamma }}$$

  • 1
    $\begingroup$ (+1) for this solid answer. Well done proposing the Lemma. I found that it's all we need. Once equipped with it, we can easily complete the proof of the generalized relationship using only integration by parts and change of variables. We end up having to show that $\gamma$ as given by $\gamma=-\int_0^\infty \log(x)e^{-x}\,dx$, is equivalent to $\gamma$ as given by the limit $\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n \frac1k\right)$. I've posted an alternative approach that references your Lemma. Let me know your thoughts. And thank you for your excellent solution! -Mark $\endgroup$ – Mark Viola Jan 17 '17 at 16:39
  • $\begingroup$ very nice and clever answer and I like it+1) $\endgroup$ – xpaul Feb 11 '17 at 1:20

I thought it might be instructive to post a solution that leverages the Lemma posted by @Renascence_5. To that end, we proceed.

The Lemma proved in the aforementioned post is expressed as

$$\int_0^\infty \frac{e^{-x^a}-\cos(x^a)}{x}\,dx=0 \tag 1$$

for $a>0$.

We now examine a generalized version of the integral of interest and write

$$I(a,b)=\int_0^\infty \frac{\cos(x^a)-\cos(x^b)}{x}\,dx \tag 2$$

for $a>0$ and $b>0$.

Using $(1)$ reveals that $(2)$ can be written as

$$I(a,b)=\int_0^\infty \frac{e^{-x^a}-e^{-x^b}}{x}\,dx \tag 3$$

Next, we integrate by parts the integral in $(3)$ with $u=e^{-x^a}-e^{-x^b}$ and $v=\log(x)$ to obtain

$$\begin{align} I(a,b)&=\int_0^\infty \left(ax^{a-1}e^{-x^a}-bx^{b-1}e^{-x^b}\right)\,\log(x)\,dx\\\\ &=\int_0^\infty ax^{a-1}e^{-x^a}\,\log(x)\,dx-\int_0^\infty bx^{b-1}e^{-x^b}\,\log(x)\,dx\\\\ &=\frac1a \int_0^\infty e^{-x}\,\log(x)\,dx-\frac1b \int_0^\infty e^{-x}\,\log(x)\,dx\\\\ &=-\left(\frac1a -\frac1b\right)\,\gamma \end{align}$$

where we used the integral relationship $\gamma =-\int_0^\infty e^{-x}\,\log(x)$.


We can show that $\gamma$ as given by $\gamma=-\int_0^\infty e^{-x}\,\log(x)\,dx$ is equal to $\gamma$ as expressed by the limit $\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)$ as follows.

$$\begin{align} \int_0^\infty e^{-x}\,\log(x)\,dx&=\lim_{n\to \infty}\int_0^n \left(1-\frac xn\right)^n\,\log(x)\,dx\\\\ &=\lim_{n\to \infty} n \int_0^1 x^n \log(n(1-x))\,dx\\\\ &=\lim_{n\to \infty} n \left(\log(n) \int_0^1 x^n\,dx+\int_0^1 x^n\,\log(1-x)\,dx\right)\\\\ &=\lim_{n\to \infty} \left(\frac{n}{n+1}\log(n)+\int_0^1 x^n\,\log(1-x)\,dx\right)\\\\ &=\lim_{n\to \infty} \left(\frac{n}{n+1}\log(n)-n\sum_{k=1}^\infty \frac1k \int_0^1 x^{n+k}\,dx\right)\\\\ &=\lim_{n\to \infty} \left(\frac{n}{n+1}\log(n)-n \sum_{k=1}^\infty \frac1{k(k+n+1)}\right)\\\\ &=\lim_{n\to \infty} \frac{n}{n+1}\left(\log(n)- \sum_{k=1}^\infty \left(\frac1k-\frac1{k+n+1}\right)\right)\\\\ &=\lim_{n\to \infty} \frac{n}{n+1}\left(\log(n)- \sum_{k=1}^{n+1} \frac1k\right)\\\\ &=\lim_{n\to \infty} \left(\log(n)- \sum_{k=1}^{n} \frac1k\right)\\\\ \end{align}$$

as was to be shown!

  • $\begingroup$ Is there a typo in your 6th-to-last line, namely $$\lim_{n\to \infty} \left(\frac{n}{n+1}\log(n)+n \int_0^1 x^n\,dx+\int_0^1 x^n\,\log(1-x)\,dx\right)$$ It looks like there is an extra $$n \int_0^1 x^n\,dx$$ $\endgroup$ – clathratus Apr 16 '19 at 18:16
  • $\begingroup$ @clathratus Indeed there was a typographical error that I have edited accordingly. Thank you for alerting me. $\endgroup$ – Mark Viola Apr 16 '19 at 18:38

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

A '$\ds{\color{#f00}{complex\ like}}$' answer can still be useful for other users. So, that is the reason I like to put forward the following answer:

\begin{align} &2\int_{0}^{\infty}{\cos\pars{x^{2}} - \cos\pars{x} \over x}\,\dd x = 2\lim_{\Lambda \to \infty}\bracks{% \int_{0}^{\Lambda}{1 - \cos\pars{x} \over x}\,\dd x - \int_{0}^{\Lambda}{1 - \cos\pars{x^{2}} \over x}\,\dd x} \\[5mm] = &\ 2\lim_{\Lambda \to \infty}\bracks{% \int_{0}^{\Lambda}{1 - \cos\pars{x} \over x}\,\dd x - {1 \over 2}\int_{0}^{\Lambda^{2}}{1 - \cos\pars{x} \over x}\,\dd x} \label{1}\tag{1} \end{align}

With $\ds{R > 0}$:

\begin{align} &\int_{0}^{R}{1 - \cos\pars{x} \over x}\,\dd x = \Re\int_{0}^{R}{1 - \expo{\ic x} \over x}\,\dd x \\[5mm] = & -\Re\int_{0}^{\pi/2}\bracks{1 - \exp\pars{\ic R\expo{\ic\theta}}}\ic\,\dd\theta - \Re\int_{R}^{0}{1 - \expo{-y} \over y}\,\dd y \\[5mm] = &\ -\int_{0}^{\pi/2}\sin\pars{R\cos\pars{\theta}} \expo{-R\sin\pars{\theta}}\,\dd\theta + \ln\pars{R}\pars{1 - \expo{-R}} - \int_{0}^{R}\ln\pars{y}\expo{-y}\,\dd y \\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\, &\ \ln\pars{R} - \int_{0}^{\infty}\ln\pars{y}\expo{-y}\,\dd y = \bbx{\ds{\ln\pars{R} + \gamma}} \label{2}\tag{2} \end{align}


$$ \left\{\begin{array}{l} \ds{0 < \verts{\int_{0}^{\pi/2}\sin\pars{R\cos\pars{\theta}} \expo{-R\sin\pars{\theta}}\,\dd\theta} < \int_{0}^{\pi/2} \exp\pars{-\,{2R \over \pi}\,\theta}\,\dd\theta = {\pi \over 2}\,{1 - \expo{-R} \over R}} \\[5mm] \mbox{and}\ \ds{\int_{0}^{\infty}\ln\pars{y}\expo{-y}\,\dd y = \left.\totald{}{\mu}\int_{0}^{\infty}y^{\mu}\expo{-y}\,\dd y\, \right\vert_{\ \mu\ =\ 0} = \left.\totald{\Gamma\pars{\mu + 1}}{\mu}\right\vert_{\ \mu\ =\ 0} = \Psi\pars{1} = -\gamma} \end{array}\right. $$

With \eqref{1} and \eqref{2}: \begin{align} 2\int_{0}^{\infty}{\cos\pars{x^{2}} - \cos\pars{x} \over x}\,\dd x & = 2\braces{\bracks{\ln\pars{\Lambda} + \gamma} - {1 \over 2}\bracks{\ln\pars{\Lambda^{2}} + \gamma}} = \bbx{\ds{\gamma}} \end{align}

  • $\begingroup$ Hi Felix. I really appreciate your writing this solution. However, in the OP, I cited a complex solution that I had already given, and this is quite similar in kind. Also, in the OP, I did state quite explicitly that I was seeking real analysis solutions only. So, unfortunately your fine solution does not qualify for this particular question. -Mark $\endgroup$ – Mark Viola Feb 14 '17 at 4:34
  • $\begingroup$ @Dr.MV Sorry. I just read the top. I didn't care about the OP further details. Hereafter, I'll start to read the 'complete question'. Anyway, I'll leave my answer over here . It's not what the OP request but it'll be useful to other users. In another hand, it's like a 'personal cloud' for my answers. Thanks for your remarks. $\endgroup$ – Felix Marin Feb 14 '17 at 20:17
  • $\begingroup$ Sure Felix. Just curious ... You do know that I am the OP here, yes? $\endgroup$ – Mark Viola Feb 14 '17 at 20:37
  • $\begingroup$ @Dr.MV Thanks. I knew you were the OP. $\endgroup$ – Felix Marin Feb 23 '17 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.