Proving differentiability Prove that $$f(t) = \int_{0}^{1} e^{tx} x^{-1/3} dx $$ is differentiable at every $t \in \mathbb{R}$. 
ATTEMPT: I show $f'(t)$ exists at every $t \in \mathbb{R}$, i.e. the limit $$\lim_{h \to 0} \int_{0}^{1} f_{h}(x;t) \ dx $$ exists, where $$f_{h}(x;t) = e^{tx}x^{-1/3} \frac{(e^{hx}-1)}{h}.$$
I want to take the limit inside the integral by using dominated convergence theorem but I am having trouble finding the right dominating function. 
 A: Let $M$ be a real number such that $M>0.$ Assume that $t \in [-M,M]$.
Each function $\displaystyle x \mapsto f_t(x):=e^{tx} x^{-1/3}$ is integrable:
$$ \int_{0}^1 f_t(x) \:dx=\int_{0}^1 e^{tx} x^{-1/3} \:dx < \int_{0}^1 e^{M}x^{-1/3} \:dx =\frac32 e^{M}<\infty$$
Now, we have $\displaystyle x \mapsto f'_t(x):=e^{tx} x^{2/3}$ which is integrable too:
$$ \int_{0}^1 f'_t(x) \:dx=\int_{0}^1 e^{tx} x^{2/3} \:dx < \int_{0}^1 e^{M}x^{2/3} \:dx =\frac35 e^{M}<\infty$$
Consequently, by the dominated convergence theorem p. 6, your initial integral is differentiable on each $[-M,M]$, thus it is differentiable on $\mathbb{R}$.
A: For concrete examples you often can do it directly by asymptotic methods and can see much more information than the dominated convergence theorem can tell you, such as the limit as well as the rate of convergence itself.
$e^{hx} \in 1 + hx + [c]h^2x^2$ for any $x \in [0,1]$ for some $c \in \mathbb{R}$ where $[c] = \{ x : |x| \le |c| \}$.
Thus $e^{tx} x^{-\frac{1}{3}} \dfrac{e^{hx}-1}{h} \in e^{tx} \left( x^{\frac{2}{3}} + [ch]x^{\frac{5}{3}} \right)$ for any $h \ne 0$.
Thus $\int_0^1 e^{tx} x^{-\frac{1}{3}} \dfrac{e^{hx}-1}{h}\ dx \in \int_0^1 e^{tx} x^{\frac{2}{3}}\ dx + \left[ ch \int_0^1 e^{tx} x^{\frac{5}{3}}\ dx \right] \to \int_0^1 e^{tx} x^{\frac{2}{3}}\ dx$ as $h \to 0$.
