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Simple question about some function:

I define $$y =\exp\left(\int^x f(t)w(t) dt \right) $$ and i want to take the derivative with respect to x. Can I just say y'= $$f(x)w(x) \exp \left(\int^x f(t) w(t) dt\right) $$?

Consequently can I say that y'' = $$ \left( f'(x)w(x)+f(x)w'(x) \right) \exp\left(\int^x f(t) w(t) dt\right) +f(x)w(x)f(x)w(x)\exp\left(\int^x f(t) w(t) dt\right)= $$ $$\left( f'(x)w(x)+f(x)w'(x) \right) \exp\left(\int^x f(t) w(t) dt\right) +f(x)^2 w(x)^2 \exp\left(\int^x f(t) w(t) dt\right) $$

Generally, what is the difference between $$\int^x f(t)w(t)dt $$ and just $$\int f(x)w(x)dx $$

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What are aaaaaaalllll the lower limits of the above integrals?! – DonAntonio Dec 2 '12 at 17:23
Is it just notation ? I dont understand it completely. I assume its just the indefinite integral with respect to x? – Applied mathematician Dec 2 '12 at 17:26
perhaps it is "just notation", but it might be important, depending on the functions involved. Anyway, I think the lower limtis are lacking. – DonAntonio Dec 2 '12 at 17:30
If your author means $$\int^xf(t)w(t)dt=\int_0^{x}f(t)w(t)dt$$ everything works out. (except for the second derivative, I think) – 000 Dec 2 '12 at 17:47
My mistake. Your second derivative is also correct! :) – 000 Dec 2 '12 at 17:56
up vote 0 down vote accepted

The definition $$x\mapsto y(x):=\exp\left(\int^x f(t)w(t) dt \right) $$ defines the function $y(\cdot)$ only up to a multiplicative constant $\ne0$. In fact, the right side has to be interpreted as follows: Assume that $t\mapsto F(t)$ is an arbitrary primitive of $t\mapsto f(t)w(t)$. Then $y(x):=\exp\bigl(F(x)\bigr)$. Two such primitives $F_1$ and $F_2$ differ by an additive constant $c\in{\mathbb R}$, therefore two candidates $x\mapsto y(x)$ satisfying your definition differ by a multiplicative constant $C:=e^c\ne 0$.

From $y(x)=\exp\bigl( F(x)\bigr)$ we get $$y'(x)=\exp\bigl( F(x)\bigr)\cdot F'(x)=y(x)\cdot f(x)w(x)$$ and then $$y''(x)=y(x)\bigl(f^2(x)w^2(x)+f'(x)w(x)+f(x)w'(x)\bigr)\ .$$ The "arbitrary multiplicative constant" is still present on the right side of the last equation, but the result is written in such a way that after this constant has been chosen once and for all no further ambiguity persists.

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Are you referring to $\int_{0}^{x}f(t)w(t)dt$ and $\int f(x)w(x)dx$? Well, if we differentiate the first and second, we arrive at $f(x)w(x)$, but for conceptually different reasons.

We know that $$\frac{d}{dx}\int_{0}^{x}f(t)w(t)dt=f(x)w(x)$$

due to the following proof (part of the Fundamental Theorem of Calculus):

Let $u(t)=f(t)w(t)$. Define $U(t)=\int u(t)dt$. From this, we have $$ \begin{align} \frac{d}{dx}\int_0^{x}f(t)w(t)dt&=\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\frac{d}{dx}\left[ \int u(t)dt\right]^{x}_{0}\\ &=\frac{d}{dx}\left[U(x)-U(0)\right]\\ &=\frac{d}{dx}U(x)+\frac{d}{dx}U(0)\\ &=\frac{d}{dx}U(x)+0\\ &=u(x)\\ &=f(x)w(x). \end{align} $$

We know that $$\frac{d}{dx}\int f(x)w(x)dx=f(x)w(x)$$ on the basis of the fact that differentiation and integration are inverse operations just like addition and subtraction. They cancel one another, in other words.

Attempt To Address Your Concerns More Aptly

Let's assume the author means $$y=\exp\left(\int_0^{x}f(t)w(t)dt\right).$$

Make the $u$ sub as above and say more aptly $$y=\exp\left(\int_0^{x}u(t)dt\right).$$

When we differentiate this, we must apply the chain rule: $$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}.$$

Let $v=\int_0^{x}u(t)dt$. We then have:

$$ \begin{align} \frac{dy}{dx}&=\exp(v)\frac{d}{dx}v\\ &=\exp\left({\int_0^x u(t)dt}\right)\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\exp\left({\int_0^x u(t)dt}\right)u(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f(x)w(x). \end{align}$$

Second Derivative

$$ \begin{align} y''&=\left[\exp\left({\int_0^x u(t)dt}\right)u(x) \right]'\\ &=\left(\exp\left({\int_0^x u(t)dt}\right)\right)'u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x) \quad (\text{product rule: } (fg)'=f'g+fg'.)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u(x)u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)w(t))'\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)'w(t)+f(t)w'(t)). \end{align} $$

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When one writes $$ \int g(x)dx, $$ the usual meaning is "an antiderivative of g".

When one writes $$ \int_a^bg(x)dx, $$ one means the "integral of g on $[a,b]$" (or "definite integral"), which is a number defined using a limit of Riemann sums.

If one writes $$ \int_a^xg(t)dt, $$ then for each $x$ this gives some number (i.e., for each $x$ one gets a definite integral, that is a number) and so we have a function on $F(x)$. The Fundamental Theorem of Calculus states the fact that in this last case $F(x)$ is an antiderivative for $f(x)$.

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and what does $$\int^x f(t) dt $$ mean? is it the same as F(X) ? – Applied mathematician Dec 2 '12 at 17:37
Probably. But it's a very unusual notation. – Martin Argerami Dec 2 '12 at 17:46

If the functions $\,f,w\,$ are such that $\,f(t)w(t)\,$ has a primitive $\,F(t)\,$ in some interval containing $\,x\,$ , say in $\,[a,x]\,$ , then the FTC tells us that

$$\int_a^xf(t)w(t)\,dt=F(x)-F(a)$$ so

$$y=e^{\int_a^xf(t)w(t)\,dt=F(x)-F(a)}\Longrightarrow \frac{dy}{dx}=y'=F'(x)\,e^{F(x)-F(a)}=f(x)w(x)\,e^{F(x)-F(a)}$$

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