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}