Hint: You are not looking for $\min(p,q)$ but for a term that looks like:
$$|p+q|-|p-q|=2\min(p,q)$$
There is a similar formula for $\max(p,q)$
$$|p+q|+|p-q|=2\max(p,q)$$
Also consider that we can write
$$\sin(px)\sin(qx) = \dfrac{\cos((p-q)x)-\cos((p+q)x)}{2}$$
I think you can see a clear resemblance in the formulae.
It is another important note that $$\int_0^{\infty} \dfrac{\cos(tx)}{x^2} dx$$ does not converge because of issues around zero. Instead consider $$\int_0^{\infty} \dfrac{\cos(ax)-\cos(bx)}{x^2} dx$$
As shown by @eyeballfrog,
$$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx =\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du$$
Further manipulating:
$$\begin{align}
\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du
&=\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\
&=\frac12\int_{-\pi}^\pi \frac{\sin[\sin(u)]}{\sin(u)}\left(\frac{1-\cos u}{2}\right)e^{-\cos(u)}du \\
&=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}e^{-\cos(u)+i\sin u}du \\
&=\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{-iu})du \\
&=-\frac14\Im\int_{-\pi}^\pi \frac{1-\cos u}{\sin(u)}\exp(-e^{iu})du \\
\end{align}
$$
Let $z=e^{iu}$, then
$$\begin{align}
\int_0^\pi \frac{\sin[\sin(u)]}{\sin(u)}\sin^2\left(\frac{u}{2}\right)e^{-\cos(u)}du
&=-\frac14\Im\int_{-\pi}^\pi \frac{2-2\cos u}{2\sin(u)}\exp(-e^{iu})du \\
&=-\frac14\Im\oint_{|z|=1} \frac{2-z-z^{-1}}{(z-z^{-1})/i}\exp(-z)\frac{dz}{iz} \\
&=\frac14\Im\oint_{|z|=1} \frac{z^2-2z+1}{z^2-1}\frac{e^{-z}}z dz\\
&=\frac14\Im\underbrace{\oint_{|z|=1} \frac{z-1}{z+1}\frac{e^{-z}}z dz}_{I}\\
\end{align}
$$
Note that the contour integral has to be understood in the Cauchy principal value sense, since a pole lies on the path of integration.
Consider the contour $C$, a unit circle with a semicircle indent to the right at $z=-1$.
By residue theorem, $$\oint_C \frac{z-1}{z+1}\frac{e^{-z}}z dz=2\pi i\operatorname*{Res}_{z=0}\frac{z-1}{z+1}\frac{e^{-z}}z$$
$$\implies I+\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-2\pi i$$
Since the indent is a semicircle and goes clockwisely, it is not difficult to prove that
$$\int_{\text{indent}}\frac{z-1}{z+1}\frac{e^{-z}}z dz=-\frac12\cdot 2\pi i\operatorname*{Res}_{z=-1}\frac{z-1}{z+1}\frac{e^{-z}}z=-2\pi i\cdot e$$
Hence, $$I=2\pi i (e-1)$$
As a result,
$$\int_0^{\infty} \frac{e^{\cos(x)}\sin(\sin(x))}{x} dx=\frac\pi 2(e-1)$$
which has been confirmed numerically.
Best Answer
On a little semicircle $\;C_\epsilon\;$ (upper) of radius $\;\epsilon>0\;$ about zero, you indeed have to take the limit $\;\epsilon\to0\;$ and then you get by the corollary to the lemma here , that
$$\lim_{\epsilon\to0}\int_{C_\epsilon}\frac{(z+1)^2}{z(z^2+4)^2}dz=\pi i\,\text{Res}\,(f)_{z=0}=\frac{\pi i}{16}$$
The rest is standard: the upper semicircle of radius $\;R>>0\;$ so that you only need the poles with positive imaginary part, take residues, limits and etc.
Added on request: The actual "problem" is to deal with the integral on the arc $\;\gamma_R:=\{z=Re^{it}\;/\;0<t<\pi\}\;$ , but either the L-M (Estimmation Lemma) or Jordan's Lemma solve that and this part of the integral, in this case, tends to zero when $\;R\to\infty\;$.
Also, the residue at $\;z=2i\;$, which is a double pole (observe that we don't care about the other pole $\;z=-2i\;$, as we are on the upper semiplane) is
$$\text{res}\,(f)_{z=2i}=\lim_{z\to 2i}\left((z-2i)^2\frac{(z+1)^2}{z(z^2+4)^2}\right)'=\lim_{z\to2i}\left(\frac{(z+1)^2}{z(z+2i)^2}\right)'$$$${}$$
$$=\lim_{z\to2i}\frac{2z(z+1)(z+2i)-(z+1)^2\left((z+2i)+2z\right)}{z^2(z+2i)^3}=\frac{4i(1+2i)\cdot4i-(1+2i)^2(4i+4i)}{(-4)(-64i)}=$$$${}$$
$$=\frac{-16-32i-(-3+4i)\cdot8i}{256i}=\frac{-16-32i+24i+32}{256i}=-\frac1{32}-\frac i{16}$$
so we get, calling $\;C\;$ the whole simple, closed contour:
$$-\frac1{32}-\frac i{16}=\oint_{C}f(z)dz=\int_{-R}^{-\epsilon}f(x)dx-\int_{\gamma_\epsilon}f(z)dz+\int_\epsilon^Rf(x)dx+\int_{\gamma_R}f(z)dz$$
Observe the minus sign before the integral around zero because when we go from $\;-R\to R\;$ that half semicircle's integration is done in the negative direction! Well, now just take the double limit and use Cauchy's Theorem:
$$\lim_{R\to\infty,\,\epsilon\to0}\left(-\frac1{32}-\frac i{16}\right)=\frac1{2\pi i}\lim_{R\to\infty,\,\epsilon\to0}\int_Cf(z)dz=\frac1{2\pi i}\left(\int_{-\infty}^\infty f(x)dx-\frac{\pi i}{16}\right)\implies$$$${}$$
$$\implies\int_{-\infty}^\infty\frac{(x+1)^2}{x(x^2+4)^2}dx=\frac\pi8 $$