Finding $\int\limits_1^\infty \frac{\sin^4(\log x)}{x^2 \log x} \mathrm{d}x$ How do we prove that $$\int_{1}^{\infty} \frac{\sin^4(\log x)}{x^2 \log x} \mathrm{d}x=\dfrac{\log\left(\dfrac{625}{17}\right)}{16}$$
I tried substitutions like $\log x=\arcsin t$, but it doesn't seem to work out. Please help me out. Thank you.
 A: The integral screams for a sub $x=e^u$; the result is
$$\int_0^{\infty} du \, e^{-u} \frac{\sin^4{u}}{u} $$
This is very computable by introducing a parameter and differentiating under the integral.  In this case, consider
$$F(k) = \int_0^{\infty} du \, e^{-k u} \frac{\sin^4{u}}{u} $$
$$F'(k) = -\int_0^{\infty} du \, e^{-k u} \sin^4{u} $$
$F'(k)$ is relatively easy to compute using the fact that $\sin^4{u} = \frac{3}{8}-\frac12 \cos{2 u} + \frac18 \cos{4 u}$, and that
$$\int_0^{\infty} du \, e^{-k u} \cos{m u} =  \frac{k}{k^2+m^2}$$
Thus
$$F'(k) = -\frac{3}{8 k} + \frac12 \frac{k}{k^2+4} - \frac18 \frac{k}{k^2+16} $$
and
$$F(k) = -\frac1{16} \log{\left [  \frac{k^6 (k^2+16)}{(k^2+4)^4} \right ]} + C$$
To evaluate $C$, we must consider $\lim_{k \to \infty} F(k)$ because $F(0)$ represents a non convergent integral.  Because the limit is zero, we must have $C=0$.  The integral we seek is then
$$F(1) = \frac1{16} \log{\frac{625}{17}}$$
A: I propose the following approach:


*

*Let $t=\dfrac1x$

*Let $u=\ln t$.

*Use the power-reduction formula for $\sin^4x$.

*Employ Euler's formula in conjunction with the linear properties of the integral.

*Express your new integral(s) in terms of $I(k)=\displaystyle\int_0^\infty e^{-kx}~dx$. You'll have $\displaystyle\int I(k)~dk$, where $I(k)$ will turn out to be a rational function in k. Solve it by using the usual methods.

