Integral of a function changing sign By some numerical tests, we can see that the following function is negative on $(0,1)$:
$$\small f(x)=\int_0^\infty\frac{s^{x-1} e^{-2 s} (\pi \cos(\pi x) (s^{2 x}+(0.1)^2)-\sin(\pi x) \ln(s) (s^{2 x}-(0.1)^2)+2 \pi (0.1) s^x )}{(2 (0.1) s^x \cos(\pi x)+s^{2 x}+(0.1)^2)^2} ds.$$
Note that the integrated function change sign for specific values of $x$.
Is there any technique to show that such an integral is negative?
 A: In fact, this conjecture fails to hold for $x$ in a right neighborhood of $0$.
Indeed, let $g(x,s)$ denote the integrand. Note that $g(x,s)$ is continuous in $x\in[0,1)$ for each real $s>0$, and
$$g(0+,s)=g(0,s)=\frac{100 \pi  e^{-2 s}}{121 s}$$
for all real $s>0$. So, by the Fatou lemma,
$$f(0+)\ge\int_0^\infty g(0+,s)\,ds=\infty>0,$$
which contradicts the conjecture that $f<0$ on $(0,1)$.

The above reasoning is not quite correct -- because the functions $g(x,\cdot)$ with $x\in(0,1)$ do not seem to be all bounded from below by an integrable function, and therefore the Fatou lemma cannot be applied.
The answer still remains negative, though, but we have to work harder to get it.
In what follows, $x\in(0,1/2)$ and $s\in(0,\infty)$, so that $\cos\pi x>0$ and $\sin\pi x>0$.
Removing manifestly positive terms in the numerator of the ratio expressing $g(x,s)$, we get
\begin{equation}
    \frac{g(x,s)}{\sin\pi x}\ge g_1(x,s):=g_2(x,s)e^{-2s}, \tag{1}\label{1}
\end{equation}
where
\begin{equation}
g_2(x,s)    :=
    \frac{s^{x-1} (s^{2 x}-1/100) \ln s}{(c s^x/5+s^{2 x}+1/100)^2},\quad c:=\cos\pi x>0.  \tag{2}\label{2}
\end{equation}
So,
\begin{equation}
\frac{f(x)}{\sin\pi x}\ge\int_0^\infty g_1(x,s)\,ds \tag{3}\label{3}
\ge f_1(x):=\int_0^1 g_1(x,s)\,ds,
\end{equation}
since $g_1(x,s)\ge0$ if $s\ge1$. In what follows, $x\in(0,1/2)$ and $s\in(0,1)$.
Next,
\begin{equation}
    \int_0^1 g_2(x,s)\,ds=
    10\frac{ \tan ^{-1}\frac{c+10}{\sqrt{1-c^2}}-
    \tan^{-1}\frac{c}{\sqrt{1-c^2}}}{x^2\sqrt{1-c^2}}
    \sim\frac{100}{11x^2} \tag{4}\label{4}
\end{equation}
as $x\downarrow0$.
By \eqref{1} and \eqref{2}, $|g_1(x,s)-g_2(x,s)|\le2s|g_2(x,s)|\le2\times10^4|\ln s|$. Since $\int_0^1 |\ln s|\,ds<\infty$, it follows from \eqref{3} and \eqref{4} that
\begin{equation}
\frac{f(x)}{\sin\pi x}\ge\frac{100-o(1)}{11x^2}
\end{equation}
as $x\downarrow0$, which implies that $f>0$ in a right neighborhood of $0$. $\quad\Box$
