For $t \in \mathbb{R}$, let $$ I_1(t) = \int_{0}^1 \frac{x e^x}{1 + \cosh(t x)} dx $$ and $$ I_2(t) = \int_{0}^\infty \frac{x e^{-x}}{1 + \cosh(t x)} dx. $$
I want to show $I_1(t) \geq I_2(t)$ with equality only if $t = 0$. Numerical integration suggests this is true. Any suggestions?
This is only a partial answer:
We have $I_1 (0) = I_2 (0) = \frac{1}{2}$ and since $I_1$ and $I_2$ are symmetric functions we can assume $t > 0$ from now on. Then \begin{align} I_1 (t) &= \int \limits_0^1 \frac{x \mathrm{e}^{x}}{1 + \cosh(tx)} \, \mathrm{d} x \stackrel{t x = 2 u}{=} \frac{2}{t^2} \int \limits_0^{t/2} \frac{u \mathrm{e}^{2u/t}}{\cosh^2(u)} \, \mathrm{d} u \, , \\ I_2 (t) &= \int \limits_0^\infty \frac{x \mathrm{e}^{-x}}{1 + \cosh(tx)} \, \mathrm{d} x \stackrel{t x = 2 u}{=} \frac{2}{t^2} \int \limits_0^\infty \frac{u \mathrm{e}^{-2u/t}}{\cosh^2(u)} \, \mathrm{d} u \, . \end{align} Using Taylor series and a few known integrals we can derive the asymptotic behaviour of the integrals (working with the $x$-integrals for small and the $u$-integrals for large values of $t$, of course): \begin{align} I_1(t) &\sim \begin{cases}\frac{1}{2} \left[1 - \frac{3 - \mathrm{e}}{2} t^2 + \mathcal{O} (t^4)\right] , & t \to 0^+ \\ \frac{2 \log(2)}{t^2} \left[1 + \frac{\pi^2}{6 \log(2) t} + \mathcal{O} \left(\frac{1}{t^2}\right)\right], & t \to \infty\end{cases}, \\ I_2(t) &\sim \begin{cases}\frac{1}{2} \left[1 - \frac{3}{2} t^2 + \mathcal{O} (t^4)\right] , & t \to 0^+ \\ \frac{2 \log(2)}{t^2} \left[1 - \frac{\pi^2}{6 \log(2) t} + \mathcal{O} \left(\frac{1}{t^2}\right)\right], & t \to \infty\end{cases}. \end{align} These results show that $I_1(t) > I_2(t)$ holds for sufficiently small or large $t$.
In order to obtain the inequality on a larger interval we write \begin{align} I_2 (t) &= 2 \int \limits_0^\infty x \mathrm{e}^{-x} \frac{\mathrm{e}^{-t x}}{(1 + \mathrm{e}^{-t x})^2} \, \mathrm{d} x \stackrel{\text{IBP}}{=} \frac{2}{t} \int \limits_0^\infty \frac{(x-1) \mathrm{e}^{-x}}{1 + \mathrm{e}^{-tx}} \, \mathrm{d}x \\ &= \frac{2}{t} \sum \limits_{k=0}^\infty (-1)^k \int \limits_0^\infty (x-1) \mathrm{e}^{-(1+kt)x} \, \mathrm{d} x = 2 \sum \limits_{k=1}^\infty \frac{(-1)^{k-1} k}{(1+kt)^2} \end{align} (interchanging integration and summation is justified by the dominated convergence theorem). Doing the same for $I_1$ we find the series representation $$ I_1(t) - I_2(t) = \frac{2}{t} \sum \limits_{k=1}^\infty (-1)^{k-1} f(kt) \, , \tag{1}$$ where $f \colon \mathbb{R}^+ \to \mathbb{R}^+$ is defined by $$ f(r) = r \left(\frac{1 - r \mathrm{e}^{1-r}}{(1-r)^2} - \frac{1}{(1+r)^2}\right) \, . $$ The apparent singularity at $r=1$ is removed by setting $f(1) = \frac{1}{4}$. $f$ is indeed a positive function, since letting $r = \frac{1 - s}{1 + s}$ and rearranging the inequality shows that $$ f(r) > 0 ~ \forall ~ r \in (0,\infty) \setminus \{1\} ~ \Leftrightarrow ~ \log(1+s) > \frac{s}{1+s} ~ \forall ~ s \in (-1,1) \setminus \{0\}$$ and this elementary inequality is true. Moreover, taking the derivative we see that $f$ increases from $f(0^+) = 0$ to a maximum at $r_0 \simeq 1.8143$ before monotonically approaching zero again. In particular, the sequence $(f(kt))_{k \in \mathbb{N}}$ is strictly decreasing to zero if $t \geq r_0$ (and, in fact, for some smaller $t$ as well). But this implies that the alternating series in equation $(1)$ is positive, so $I_1(t) > I_2(t)$ holds for $t \in [r_0,\infty)$.
This method does not work on the remaining intervall $(0,r_0)$, however (ironically, this is where $I_1(t) - I_2(t)$ is largest numerically). At the moment, I can only think of piecing together a few Taylor expansions and their error estimates in this region. We have, for example ($\operatorname{Li}_2$ is the dilogarithm), \begin{align} I_1 (1) &= \frac{\pi^2}{6} - 2 \log(2) + 2 \left[2 \log(1 + \mathrm{e}) + \operatorname{Li}_2 \left(-\mathrm{e}\right) - \frac{\mathrm{e}}{1 + \mathrm{e}}\right] \, , \\ I_2 (1) &= \frac{\pi^2}{6} - 2 \log(2) \end{align} and the expression in the square brackets is positive, so the inequality holds in some neighbourhood $(\varepsilon, 2 -\varepsilon)$ of $1$. It should be possible (but probably tedious) to make $\varepsilon > 0$ small using Taylor polynomials. Combining this with the previous results would then complete the proof, but this is clearly not a very elegant method.
