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i'm using casella and burger pg. 82 problem 2.39 a) where it says "In each of the following calculate the indicated derivatives justifying all operations".
\begin{equation} \frac{d}{dx} \int_{0}^{x} e^{-\lambda t} dt \end{equation}

there is a theorem built from Leibnitz's theorem which states that if the function f is differentiable at $\theta_{0} \forall x$ and if there is a majorizing function g such that $\frac{d}{d \theta} f(x,\theta) \leq g(x,\theta_{0})$ and g must be bounded then

\begin{equation} \frac{d}{d\theta} \int_{a(\theta)}^{b(\theta)}f(x,\theta)dx = \int_{a(\theta)}^{b(\theta)} \frac{ \partial f}{\partial \theta}(x,\theta) dx \end{equation}

if you divide $\lambda$ by $\delta_{0} < \lambda$ you can find the majorizing function. however I'm not sure how to evaluate the partial derivative after interchanging the derivative and integral when the derivative is dependent on the integral becoming evaluated.

This link goes through Leibnitz rules for interchanging.
http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

so one way to solve this is just to evaluate the integral, and then take the derivative (duh). but how can we solve this by interchanging the derivative when the integrand is a variable of t and not x, thus if you interchange, that will result in 0 (?).

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  • $\begingroup$ I think it would be useful to state the problem exactly or the source of the book $\endgroup$ – knk Jan 24 '17 at 4:52
  • $\begingroup$ I think one reason why you can not directly interchange the derivative is because the integrand is a function f(t,$\lambda$) and not x until after you evaluate, hence you can not really use Leibnitz rules ? $\endgroup$ – sophie-germain Jan 24 '17 at 5:07
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We have $$\frac{d}{dx}\int_0^x e^{-\lambda t}dt = e^{-\lambda x}$$ by the fundamental theorem of calculus.

In the situation where the dependence on $x$ is in the integrand rather than the bounds, we can sometimes take the derivative inside the integral like $$\frac{d}{dx}\int_a^bf(x,t)dt =\int_a^b\frac{\partial}{\partial x}f(x,t)dt. $$ Whether this is legal or not requires some technical conditions like the majorizing rule you brought up (related to the dominated convergence theorem).

In the situation where both the bounds, and the integrand depend on $x,$ provided that you have some technical condition like your majorizing rule, we have $$ \frac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)dt = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt $$

Note this is just applying the fundamental theorem of calculus along with the chain rule at each endpoint and then bring the derivative inside (which is the thing that requires the technical conditions).

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  • $\begingroup$ assuming f is increasing.... $\endgroup$ – sophie-germain Jan 24 '17 at 5:19
  • $\begingroup$ @arcolombo Confused... where is that assumption needed? (And increasing in what?) $\endgroup$ – spaceisdarkgreen Jan 24 '17 at 5:48
  • $\begingroup$ sorry.pedantic. if f is decreasing then the limits (end points) would be reversed such that $f(x,a(x))a'(x)- f(x,b(x))b'(x) + \int \frac{\partial}{\partial x}f(x,t)dt$ ? do we switch the boundary points if f is decreasing function? $\endgroup$ – sophie-germain Jan 25 '17 at 18:19
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    $\begingroup$ @arcolombo no... the limits aren't reversed somehow if the $f$ is decreasing. Think of integrating a nearly horizontal line that angles slightly up vs one that angles slightly down. Does anything change? $\endgroup$ – spaceisdarkgreen Jan 25 '17 at 18:53
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    $\begingroup$ @arcolombo I believe the formula is self correcting in cases like when $a(x)>b(x)$ (which is maybe what you mean?) $\endgroup$ – spaceisdarkgreen Jan 25 '17 at 18:55

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