How to calculate
$$\lim_{y\rightarrow +\infty}\left( \int_0^y \frac {\sin^4(x)}{x} \mathrm{d}x - \frac38 \ln y\right).$$
In my view,there is $$\int_0^y \frac{\sin^4 x}{x} \mathrm{d} x =\int_0^y\frac{ \cos 4 x - 4 \cos 2 x + 3}{8 x}\mathrm{d} x.$$ I try using the Taylor expansion of cosine functions and logarithmic functions to calculate this limit，but this way seem to be not work. I think there is another way to calculate it, whereas I haven't figured it out yet.
Writing $\int_0^y=\int_0^1+\int_1^y$ and $\ln y=\int_1^y\frac{dx}{x}$, we see that the limit equals $$\int_0^1\frac{\cos 4x-4\cos 2x+3}{8x}\,dx+\int_1^\infty\frac{\cos 4x-4\cos 2x}{8x}\,dx=\frac{f(2)}{2}-\frac{f(4)}{8}$$ where, for $a>0$, $$f(a)=\int_0^1\frac{1-\cos ax}{x}\,dx-\int_1^\infty\frac{\cos ax}{x}\,dx=\int_0^a\frac{1-\cos t}{t}\,dt-\int_a^\infty\frac{\cos t}{t}\,dt.$$
Writing $\int_0^a=\int_0^1+\int_1^a$ and $\int_a^\infty=\int_1^\infty-\int_1^a$, we see that $f(a)=f(1)+\ln a$.
Finally, it is known that $f(1)=\gamma$ equals the Euler–Mascheroni constant; a possible way to show this is to use the integral with $e^{-t}$ in place of $\cos t$, and the fact that $\int_0^\infty\frac{\cos t-e^{-t}}{t}\,dt=0$.
Put together, we get the answer $$\lim_{y\to+\infty}\left(\int_0^y\frac{\sin^4 x}{x}\,dx-\frac38\ln y\right)=\frac{3\gamma+2\ln2}8.$$