# Differentiation under the integral sign with $h(x,t) = e^{xt}f(t)$

I'm working on the following problem and got stuck:

Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ such that for every fixed $$x\in(-1,1)$$ the function $$g_x(t):= e^{tx}f(t) \in L^1(\mathbb{R})$$. Let $$\varphi:(-1,1)\rightarrow \mathbb{R}$$ be defined as

$$\varphi(x) = \int{e^{tx}f(t)}dt$$

Show that $$\varphi$$ is differentiable.

My attempt:

Let $$h(x,t):(-1,1)\times\mathbb{R} \longrightarrow \mathbb{R}$$ $$h(x,t) = e^{tx}f(t)$$

Then, in order to differentiate under the integral sign, we need:

$$\bullet\space h(.,t)=e^{tx}f(t)\in L^1(\mathbb{R})$$ (which we know by definition)

$$\bullet\space h(x,.)\in C(\mathbb{R}) \space\space\space[\frac{d}{dx}\left(e^{xt}f(t)\right)=t\cdot e^{xt}f(t)]$$

and

$$\bullet\space \lvert\frac{d}{dx} h(x,t)\rvert=\lvert t\cdot e^{xt}f(t)\rvert < s(t)\space$$ for some $$s(t)\in L^1(\mathbb{R})$$

This last step is where I'm stuck. I can't find such $$s(t)$$ that dominated my function. I think that if I'm able to do so, then I'd be done. Or is my approach incorrect? Thanks.