Can we compute $\int_1^{\infty} \frac{\sin ^2(\ln x)}{x^2 \ln ^2 x} d x$ without Feynman’s trick?

Question

When I came across the integral

$$I=\int_1^{\infty} \frac{\sin ^2(\ln x)}{x^2 \ln ^2 x} d x,$$

I immediately thought of the Feynman’s trick after the substitution $x\mapsto \frac{1}{x}$.

$$ I=\int_0^1 \frac{\sin ^2(\ln x)}{\ln ^2 x} d x $$

Considering the parameterised integral

$$ I(a)=\int_0^1 \frac{\sin ^2(a \ln x)}{\ln ^2 x} d x $$ Differentiating $I(a)$ w.r.t. $a$ twice, we get

$$ I^{\prime}(a)=\int_0^1 \frac{\sin (2a \ln x)}{\ln x}dx $$ and $$ I^{\prime \prime}(a) =2 \int_0^1 \cos (2 a \ln x) d x =\frac{2}{4 a^2+1} $$ Integrating back yields $$ I ^{\prime}(a)=I^{\prime}(a)-I^{\prime}(0)= \int_0^a \frac{2}{4 t^2+1} d t =\tan ^{-1}(2 a) $$ Further integration gives $$ \boxed{\int_0^1 \frac{\sin ^2(a \ln x)}{\ln ^2 x} d x =\int_0^a \tan ^{-1}(2 t) d t=a \tan ^{-1}(2 a)-\frac{1}{4} \ln \left(4 a^2+1\right)} $$ Now we can conclude that $$\int_1^{\infty} \frac{\sin ^2(\ln x)}{x^2 \ln ^2 x} d x =I(1)= \tan ^{-1} 2-\frac{\ln 5}{4} $$

My question:

Can we compute $$\int_1^{\infty} \frac{\sin ^2(\ln x)}{x^2 \ln ^2 x} d x$$ without Feynman’s trick?

Your comments and alternative methods are highly appreciated.

Answer 1

Using the following property of Laplace Transform $$ \int_0^\infty (\mathcal Lf)(t)g(t)\mathrm dt=\int_0^\infty f(s)(\mathcal Lg)(s)\mathrm ds, \qquad\mathcal{L}\{t\}(s)=\frac1{s^2} ,$$ one has \begin{eqnarray} I&=&\int_1^{\infty} \frac{\sin ^2(\ln x)}{x^2 \ln ^2 x}\mathrm d x\overset{s=\ln x}{=}\int_0^\infty\frac{e^{-s}\sin^2 s}{s^2}\mathrm ds\\ &=&\int_0^\infty e^{-s}\sin^2 s\mathcal L\{t\}(s)\mathrm ds\\ &=&\int_0^\infty \mathcal{L}\{e^{-s}\sin^2 s\}(t)t\mathrm dt\\ &=&\int_0^\infty\frac{2t}{(t+1)(t^2+2t+5)}\mathrm dt\\ &=&\int_1^\infty\frac{2(t-1)}{t(t^2+4)}\mathrm dt\\ &=&\arctan(2)-\frac14\ln5. \end{eqnarray}

Answer 2

You have shown (via $x\mapsto \frac{1}{x}$) that $\displaystyle I=\int_0^1 \frac{\sin ^2(\ln x)}{\ln ^2 x} d x$

Now, observe that $\displaystyle \frac{\sin(\ln(x)) }{\ln{x}} = \int_0^1 \cos(t \ln{x}) \, \mathrm dt$

Then setting it up as a triple integral, we have:

$$\begin{aligned} I & := \int_0^1\frac{\sin^2(\ln(x)) }{\ln^2{x}}\, \mathrm dx \\&= \int_0^1 \int_0^1 \int_0^1 \cos(t\ln{x}) \cos(y\ln{x})\;{\mathrm dy}\;{\mathrm dt}\;{\mathrm dx} \\& = \int_0^1 \int_0^1 \int_0^1 \cos(t\ln{x}) \cos(y\ln{x})\;{\mathrm dx}\;{\mathrm dt}\;{\mathrm dy} \\&= \int_0^1 \int_0^1 \frac{1 + t^2 + y^2}{t^4 - 2 t^2 (-1 + y^2) + (1 + y^2)^2} \ \;{\mathrm dt} \;{\mathrm dy} \\& = \int_0^1 \frac{1}{2} \tan^{-1}\left(\frac{2}{y^2}\right) \;{\mathrm dy} \\& = \tan^{-1}(2) - \frac{\log(5)}{4} . \end{aligned}$$

You can fill in the details.

Answer 3

Making the substitution $t = \ln(x)$ and then integrating by parts, we have

$$\int_{1}^{\infty} \frac{\sin^{2}(\ln x)}{x^{2} \ln^{2}(x)} \, \mathrm dx = \int_{0}^{\infty} \frac{e^{-t}\sin(2t)}{t} \, \mathrm dt - \int_{0}^{\infty} \frac{e^{-t} \sin^{2}(t)}{t} \, \mathrm dt.$$

Using the integral formula $$\int_{0}^{\infty} e^{-at} \sin(bt) t^{s-1} \, \mathrm dt = \frac{\Gamma(s)}{\left(a^{2}+b^{2} \right)^{s/2}} \, \sin \left(s \arctan \left(\frac{b}{a} \right) \right), $$ the value of the first integral is $$ \begin{align} \int_{0}^{\infty} \frac{e^{-t}\sin(2t)}{t} \, \mathrm dt &= \lim_{s \to 0^{+}} \int_{0}^{\infty} e^{-t} \sin(2t) t^{s-1} \, \mathrm dt \\ &= \lim_{s \to 0^{+}} \frac{\Gamma(s)}{5^{s/2}} \, \sin\left( s \arctan 2 \right) \\ &= \lim_{s \to 0^{+}} 5^{-s/2}\left(\frac{1}{s} + O(1) \right)\sin\left( s \arctan 2 \right) \\ &= \lim_{s \to 0^{+}} \frac{5^{-s/2} \sin \left(s \arctan 2 \right)}{s} \\ &=\arctan(2).\end{align}$$

And using the companion formula $$ \int_{0}^{\infty} e^{-at} \cos(bt) t^{s-1} \, \mathrm dt = \frac{\Gamma(s)}{\left(a^{2}+b^{2} \right)^{s/2}} \, \cos \left(s \arctan \left(\frac{b}{a} \right) \right), $$ the value of the second integral is $$ \begin{align} \int_{0}^{\infty} \frac{e^{-t} \sin^{2}(t)}{t} \, \mathrm dt &= \frac{1}{2} \int_{0}^{\infty} \frac{e^{-t}\left(1- \cos(2t)\right)}{t} \, \mathrm dt \\ &= \frac{1}{2}\lim_{s \to 0^{+}} \int_{0}^{\infty} e^{-t} \left(1- \cos(2t) \right) t^{s-1} \, \mathrm dt \\ &= \frac{1}{2} \lim_{s \to 0^{+}} \left(\int_{0}^{\infty} e^{-t} t^{s-1} \, \mathrm dt - \int_{0}^{\infty} e^{-t} \cos(2t) t^{s-1} \, \mathrm dt\right)\\ &= \frac{1}{2} \lim_{s \to 0^{+}} \left(\Gamma(s) - \frac{\Gamma(s)}{5^{s/2}} \cos \left(s \arctan 2 \right)\right) \\ &= \frac{1}{2}\lim_{s \to 0^{+}} \Gamma(s) \left(1- \frac{\cos \left(s \arctan 2\right)}{5^{s/2}} \right)\\ &= \frac{1}{2}\lim_{s \to 0^{+}} \left(\frac{1}{s} + O(1) \right)\left(1- \frac{\cos \left(s \arctan 2\right)}{5^{s/2}} \right) \\ &= \frac{1}{2} \lim_{s \to 0^{+}} \frac{1- 5^{-s/2}\cos \left(s \arctan 2 \right)}{s} \\ &= \frac{\ln(5)}{4}. \end{align}$$

Therefore, $$\int_{1}^{\infty} \frac{\sin^{2}(\ln x)}{x^{2} \ln^{2}(x)} \, \mathrm dx = \arctan(2) - \frac{\ln (5)}{4}. $$

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Justification for bringing the limits inside the integrals comes from the dominated convergence theorem.

For $s \in [0, \alpha], \alpha>0 $, the function $e^{-t} \sin(2t) t^{s-1}$ is dominated on the interval $(0,1]$ by the integrable function $\frac{e^{-t}|\sin(2t)|}{t}$ and dominated on the interval $(1, \infty)$ by the integrable function $e^{-t} t^{\alpha-1}$.

Similarly, the function $e^{-t} \left(1- \cos(2t) \right) t^{s-1} $is dominated on the interval $(0, 1]$ by the integrable function $\frac{e^{-t}|1-\cos(2t)|}{t} $ and dominated on the interval $(1, \infty)$ by the integrable function $2 e^{-t} t^{\alpha -1}$.

Answer 4

This might be similar to your answer, but we can use the mapping $\ln x \mapsto x$ so that

$$ I := \int_{1}^{\infty}\frac{\sin^{2}\left(\ln x\right)}{x^{2}\ln^{2}x}dx = \int_{0}^{\infty}\frac{\sin^{2}x}{x^{2}e^{x}}dx\,. $$

Using this YouTube video, we use the following Laplace transform:

$$ \mathscr{L}\left\{\frac{\sin^2t}{t}\right\}(s) = \mathscr{L}\left\{\frac{1-\cos2t}{t}\right\}(s) = \frac{1}{2}\int_{s}^{\infty}\left(\frac{1}{t}-\frac{t}{t^{2}+4}\right)dt = \frac{1}{4}\ln\left(\frac{s^{2}+4}{s^{2}}\right)\,. $$

We can use the statement

$$ \mathscr{L}\left\{f(t)\right\}(s) = F(s) \implies \mathscr{L}\left\{\frac{f(t)}{t}\right\}(s) = \int_{s}^{\infty}F\left(t\right)dt $$

and get

$$ I = \mathscr{L}\left\{\frac{\sin^2x}{x^2}\right\}(1) = \int_{1}^{\infty}\frac{1}{4}\ln\left(\frac{t^{2}+4}{t^{2}}\right)dt \overset{IBP}= \frac{1}{4}\left[t\ln\left(\frac{t^{2}+4}{t^{2}}\right)\right]_{1}^{\infty}+2\int_{1}^{\infty}\frac{1}{t^{2}+4}dt\,, $$

which simplifies down to

$$ I = \arctan2-\frac{\ln5}{4}\,. $$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{\LARGE\tt I} & \equiv \color{#44f}{\int_{1}^{\infty}{\sin^{2}\pars{\ln\pars{x}} \over x^{2} \ln^{2}\pars{x}}\dd x} \\[5mm] & = \int_{1}^{\infty}{\sin\pars{\ln\pars{x}} \over \ln\pars{x}}{\sin\pars{\ln\pars{x}} \over \ln\pars{x}}{\dd x \over x^{2}} \\[5mm] & = \Re\int_{1}^{\infty}\bracks{% {1 \over 2}\int_{-1}^{1}\expo{\ic\ln\pars{x}k} \,\,\dd k}\ \times \\ & \phantom{= \,\Re\int_{1}^{\infty}}\bracks{% {1 \over 2}\int_{-1}^{1}\expo{-\ic\ln\pars{x}q} \,\,\dd q}{\dd x \over x^{2}} \\[5mm] & = {1 \over 4}\, \Re\int_{-1}^{1}\int_{-1}^{1}\int_{1}^{\infty} x^{-2\ +\ \ic\pars{k - q}}\,\,\,\,\dd x \,\dd k\,\dd q \\[5mm] & = {1 \over 4}\, \Re\int_{-1}^{1}\int_{-1}^{1} \left.\rule{0pt}{8mm} {x^{-1 \ +\ \ic\pars{k - q}} \over -1 \ +\ \ic\pars{k - q}} \right\vert_{\,x\ =\ 1}^{\,x\ \to\ \infty}\,\,\,\,\dd x\,\dd k\,\dd q \\[5mm] & = {1 \over 4}\,\Re\int_{-1}^{1}\int_{-1}^{1} {\dd k\,\dd q \over 1 \ -\ \ic\pars{k - q}} \\[5mm] & = {1 \over 4}\int_{-1}^{1}\int_{-1}^{1} {\dd k\,\dd q \over \pars{k - q}^{2} + 1} \\[5mm] & = {1 \over 2} \int_{0}^{1}\arctan\pars{2 \over q^{2}}\dd q \\[5mm] & = \bbx{\color{#44f}{\arctan\pars{2} - {1 \over 4}\ln\pars{5}}} \approx 0.7048 \end{align}

Answer 6

Letting $s=\ln x$ transforms the integral into $$ I=\int_0^{\infty} \frac{e^{-s} \sin ^2 s}{s^2} d s $$ Then integration by parts on $\frac{1}{x^2}$ gives $$ I=\int_0^{\infty} \frac{e^{-s}\left(\sin 2 s-\sin ^2 s\right)}{s} d s $$ Noting that $e^{-s}= \int_1^{\infty} e^{-t s} d t $, we have $$ \begin{aligned} I &= \int_0^{\infty}\left[\left(\sin 2 s-\sin ^2 s\right) \int_1^{\infty} e^{-t s} d t\right] d s\\& =\int_1^{\infty} \int_0^{\infty}\left(\sin 2 s-\sin ^2 s\right) e^{-t s} d s d t \\ & =\int_1^{\infty}\left(\frac{2}{t^2+4}-\frac{2}{t(t^2+4 )}\right) d t \\ & =\left[\tan ^{-1}\left(\frac{t}{2}\right)-\frac{1}{4} \ln \left(\frac{t^2}{t^2+4}\right)\right]_1^{\infty} \\ & =\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)+\frac{1}{4} \ln \left(\frac{1}{5}\right) \\ & =\tan ^{-1} 2-\frac{\ln 5}{4} \end{aligned} $$

Answer 7

Trying to make the problem more general $$I_n=\int_1^\infty\Bigg( \frac{\sin (\log (x)) }{x\,\log (x) }\Bigg)^n\,dx=\int_0^\infty e^{-(n-1)\,t} \,\left(\frac{\sin (t)}{t}\right)^n\,dt$$

Using Euler representation of the sine function and binomial expansions, this leads to exponential integrals (have a look here for $n=2$).

The next ones are

$$I_3=\frac{5}{8} \tan ^{-1}\left(\frac{3}{2}\right)+\frac{9}{8} \cot ^{-1}(2)-\frac{3}{4} \log \left(\frac{13}{5}\right)$$ $$I_4=\frac{23}{6} \tan ^{-1}\left(\frac{2}{3}\right)-\frac{11}{12} \tan^{-1}\left(\frac{4}{3}\right)-\frac{3}{16} \log \left(\frac{1220703125}{3326427}\right)$$

Answer 8

\begin{align} &\int_1^\infty\left(\frac{\sin\ln x}{x\ln x}\right)^2dx\>\>\>\>\>\>\>\>\>\> t=\ln x\\ =&\int_0^\infty \frac{e^{-t}\sin^2 t}{t^2}\,dt =\int_0^\infty \int_0^\infty y\ e^{-(y+1)t} \sin^2 t \,dy\,dt \\ = &\int_0^\infty \frac{2y}{(y+1)[(y+1)^2+4]}\,dy=\tan^{-1}2-\frac14\ln5\\ \end{align}