Given $a\in(0,1)$ I want to prove the function \begin{align*} g(a)=\int_{\mathbb{R}}\frac{\exp(ax)}{1+\exp(x)}dx \end{align*} is infinitely differentiable.
I proved the function $f(a,x)=\frac{\exp(ax)}{1+\exp(x)}$ is integrable (with respect to $x$) for each $a\in(0,1)$, and you can clearly differentiate (with respect to $a$) to obtain $\partial_{a}f(a,x)=x\frac{\exp(ax)}{1+\exp(x)}$.
I would like to conclude by showing there is an integrable function $h$ such that for each $a\in(0,1)$ and $x\in \mathbb{R}, \; h(x)\ge |\partial_{a}f(a,x)|$ but I don't think this last condition is possible. Is there another way to prove the differentiability of $g$?
Any help would be apprecciated. Thanks!
You do not need an integrable bound which is universal for all $a\in (0,1)$. Differentiability at some $a$ in that interval only depends on the behavior of the function in some neighborhood of $a$. For any such $a$, you can consider the function $g$ on $(a-\varepsilon/2,a+\varepsilon/2)$ with $\varepsilon<a$, $a+\varepsilon<1$. Then, the function
$$ h(x)=\left\{ \begin{array}{ll} \displaystyle{x\frac{e^{(a+\varepsilon)x}}{1+e^{x}};\quad x>0}\\ \displaystyle{x\frac{e^{(a-\varepsilon)x}}{1+e^{x}};\quad x<0} \end{array} \right. $$
is an integrable upper bound for $\partial_af$ and you can apply the theorem on differentiation under the (Lebesgue) integral sign to conclude that $g$ is differentiable for $a\in(a-\varepsilon/2,a+\varepsilon/2)$ and its derivative can be found by differentiating under the integral sign. The conclusion is that $g$ is differentiable for each $a\in(0,1)$.