How to differentiate integrals with variable limits? I'd like to evaluate the following two derivatives.
$$
(1) \frac{d}{d\theta^*} \int_{\theta^{*}}^1x^{\theta}\theta^\alpha g(\theta)d\theta
$$
$$
(2) \frac{d}{dx} \int_{\theta^{*}}^1x^{\theta}\theta^\alpha g(\theta)d\theta
$$
Thank you for your time.
 A: (I will assume that $\theta^{*}$ is just a different variable unrelated to $\theta$, and that these integrals converge.)
For the first one, we have that:
$$ \frac{d}{d\theta^*} \int_{\theta^{*}}^1x^{\theta}\theta^\alpha g(\theta)d\theta = - \frac{d}{d\theta^*} \int_{1}^{\theta^{*}} x^{\theta}\theta^\alpha g(\theta)d\theta $$
By the second fundamental theorem of calculus, we have:
$$ -x^{\theta}\theta^\alpha g(\theta)$$
For the second one, we can simply interchange differentiation and integration:
$$ \int_{\theta^{*}}^1 \frac{d}{dx} x^{\theta}\theta^\alpha g(\theta) \;dx \;d\theta = \int_{\theta^{*}}^1x^{\theta-1}\theta^{\alpha+1} g(\theta)d\theta$$
A: Under certain technical assumptions:
$$\frac{\mathrm{d}}{\mathrm{d}\theta^*} \int_{\theta^*}^1 x^\theta \theta^\alpha g(\theta)\,\mathrm{d}\theta = -x^{\theta^*}\theta^{*\alpha}g(\theta^*)$$
and
$$\frac{\mathrm{d}}{\mathrm{d}x} \int_{\theta^*}^1 x^\theta \theta^\alpha g(\theta)\,\mathrm{d}\theta = \int_{\theta^*}^1 x^{\theta-1} \theta^{\alpha+1} g(\theta)\,\mathrm{d}\theta.$$
A: For number 1: Let $F(\theta)$ be an antiderivative of $x^\theta \theta^\alpha g(\theta)$ with respect to $\theta$. Then, by the second fundamental theorem of calculus,
$$\frac{\mathrm{d}}{\mathrm{d}\theta^\ast} \int_{\theta^\ast}^1 x^\theta \theta^\alpha g(\theta) \ \mathrm{d}\theta = \frac{\mathrm{d}}{\mathrm{d}\theta^\ast} \left( F(1) - F(\theta^\ast) \right)$$
Difference rule:
$$\frac{\mathrm{d}}{\mathrm{d}\theta^\ast} F(1) - \frac{\mathrm{d}}{\mathrm{d}\theta^\ast} F(\theta^\ast) $$
Since $F(1)$ is constant wrt $\theta^\ast$, and we defined $F$ conveniently, this equals
$$0 - x^{\theta^\ast} (\theta^\ast)^\alpha g(\theta^\ast) = -x^{\theta^\ast} (\theta^\ast)^\alpha g(\theta^\ast)$$
