Differentiation for a function in the integral form.

Question

I want to know generally how we differentiate a function $F(x)$ in the following form, $$F(x)=\int_a^x f(x,t)dt$$ For example, if we can work out the explicit form of $F(x)$ as the example below $$F(x)=\int_0^x e^{xt}dt=\frac{e^{x^2}-1}{x}$$ if we differentiate $F(x)$, we get $$\frac {dF(x)}{dx}=\frac{2x^2e^{x^2}-e^{x^2}+1}{x^2}$$ This simple example shows a very different approach from the case when we different $F(x)=\int_a^xf(t)dt$, which simply gives us $f(x)$.

Now suppose I have the formula for calculating the money I get after $T$ years, $P(T)$ with continuous depositing, of which the rate is given by $S(t)$, and continuous compounding, of which the annual rate is given by $r$. The formula is $$P(T)=\int_0^TS(t)e^{r(T-t)}dt$$ Since the $S(t)$ is not given explicitly, then how do I get an expression for $\frac{dP(T)}{dT}$?

I have tried using the first principle by $$\frac{dP(T)}{dT}=\lim_{\Delta T\to0} \frac{P(T+\Delta T)-P(T)}{\Delta T}$$ and I could not figure it out. Is there a way to find an expression for the derivative $\frac{dP(T)}{dT}$?

Answer 1

We have that $$ \eqalign{ & F(x + dx) - F(x) = \int_{t = a}^{x + dx} {f(x + dx,t)dt} - \int_{t = a}^x {f(x,t)dt} = \cr & = \int_{t = a}^x {\left( {f(x + dx,t) - f(x,t)} \right)dt} + \int_{t = x}^{x + dx} {f(x + dx,t)dt} = \cr & = \left( {\int_{t = a}^x {f_x (x,t)dt} } \right)dx + f(x + dx,x)dx = \cr & = \left( {\int_{t = a}^x {f_x (x,t)dt} } \right)dx + f(x,x)dx + f_x (x,x)\left( {dx} \right)^2 \quad \Rightarrow \cr }$$

$$ \Rightarrow \quad {d \over {dx}}F(x) = \int_{t = a}^x {{\partial \over {\partial x}}f(x,t)dt} + f(x,x) $$

So in your example $$ \eqalign{ & {d \over {dx}}F(x) = \int_{t = 0}^x {{\partial \over {\partial x}}e^{\,x\,t} dt} + e^{\,x^{\,2} } = \cr & = \int_{t = 0}^x {te^{\,x\,t} dt} + e^{\,x^{\,2} } = \left. {{{e^{\,x\,t} \left( {tx - 1} \right)} \over {x^2 }}\,} \right|_{t = 0}^x + e^{\,x^{\,2} } = \cr & = {{e^{\,x^{\,2} } \left( {\,x^{\,2} - 1} \right)} \over {x^2 }} + {1 \over {x^2 }} + e^{\,x^{\,2} } = {{\,2e^{\,x^{\,2} } x^{\,2} - e^{\,x^{\,2} } + 1} \over {x^2 }} \cr} $$

Answer 2

You can impose $a=Tt$ so you have that the integral is

$P(T)=\frac{1}{T}\int_0^1 S(\frac{a}{T})e^{r\frac{(T^2-a)}{T}}da$

So

$\frac{d}{dT}P(T)=\frac{1}{T}\int_0^1[-\frac{a}{T^2}S’(\frac{a}{T})e^{r\frac{(T^2-a)}{T}}+$

$+r(1+\frac{a}{T^2}) e^{r\frac{(T^2-a)}{T}} S(\frac{a}{T})]da$

Answer 3

You want to use Leibniz integral rule. If $f(x,t)$ is regular enough the following formula holds

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t)\, dt = f\big(x,b(x)\big) b'(x) - f\big(x,a(x)\big)a'(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)\, dt\,.$$