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For $a > 0$, I need to find the derivatives of $F(x) = \int\nolimits_{1}^{x}\frac{dt}{t}$ and of $G(x) = \int_{a}^{ax}\frac{dt}{t}$, $x\in \left ( 0,\infty \right )$ and use them to prove that $G(x) = F(x)$ for all $x > 0$.

I am having a hard time imaging how the final statement is true in the first place... Could someone help me imagine/picture this?

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(Travis, check out my latest edit, by clicking "[xxx] minutes ago" next to "last edited by The Chaz". It's subtle, but you might find that expressing problems without the imperative case is a more diplomatic approach in this community. Welcome, by the way!) – The Chaz 2.0 Nov 21 '11 at 22:42
@TheChaz Thank you. I should have known this first before you told me sorry. – Travis Lex Nov 22 '11 at 3:05

The first function you can compute using the Fundamental Theorem of Calculus, part II: $$\frac{d}{dx}F(x) = \frac{d}{dx}\int_1^x\frac{dt}{t} = \frac{1}{x}.$$

The second takes a bit more work; specifically, you need to use the Chain Rule; set $u=ax$. Then $$\frac{d}{dx}G(x) = \frac{d}{dx}\int_a^{ax}\frac{dt}{t} = \frac{d}{dx}\int_a^u\frac{dt}{t} = \left(\frac{d}{du}\int_a^u\frac{dt}{t}\right)\frac{du}{dx}.$$

What do you get?

If $G'(x)=F'(x)$ for all $x\gt 0$, then you know that $F$ and $G$ differ by a constant. Evaluating both at any given value of $a$ will give you that constant. How much is $F(1)$ and how much is $G(1)$?

As to why it is true, the first integral represents the area under the graph of $y = \frac{1}{t}$ from $1$ to $x$; this is the same (by one definition of the logarithm) to $\ln(x)$. The second integral is then the difference between the integral from $1$ to $ax$ and the integral from $1$ to $a$; that is, it should equal $\ln(ax) - \ln(a)$.

I would say that this is a case where calculus helps understand the picture, rather than the other way around...

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This is helpful and thorough. – The Chaz 2.0 Nov 21 '11 at 22:29

It so happens that if we are working with positive $t$, then $$\int \frac{1}{t}\,dt=\ln t +C.$$ Let's take this for granted for now. Then $$\int_a^{ax}\frac{1}{t}\,dt=\ln(ax)-\ln a.$$ But $\ln(ax)=\ln a +\ln x$. So the $\ln a$ cancel, and we get $\ln x$. The result we are after is in fact equivalent to the familiar "logarithm of a product is the sum of the logarithms."

However, in the exercise, we are not supposed to use any properties of logarithms. And we are not supposed to know a formula for the general antiderivative of $\frac{1}{t}$. Part of the reason could be that we want to formally define the natural logarithm as a certain definite integral. So we will prove the result just using the Fundamental Theorem of Calculus.

Maybe it is easiest to see what is going on by letting $H(t)$ be any antiderivative of $\frac{1}{t}$. Then $$\int_a^{ax}\frac{1}{t}\,dt=H(ax)-H(a).$$ Differentiate with respect to $x$, remembering that $a$ is to be treated as constant. We use the Chain Rule. The derivative with respect to $x$ is $aH'(ax)$. This is $a\frac{1}{ax}$, which is $\frac{1}{x}$.

So $G'(x)=\frac{1}{x}$. More easily, $F'(x)=\frac{1}{x}$. So the derivative of $G(x)-F(x)$ is $0$. This means that $G(x)$ and $F(x)$ differ by a constant. But easily $F(1)=G(1)=0$. So the constant by which $G(x)$ and $F(x)$ differ is $0$, meaning that $G(x)=F(x)$.

Another way: You were asked to use differentiation, but there are other ways to get the same result, such as substitution. Let $u=t/a$. Then $du=(1/a)dt$, and therefore $$\int_{t=a}^{ax}\frac{1}{t}\,dt=\int_{u=1}^x\frac{1}{au}a\,du=\int_{u=1}^x\frac{1}{u}du.$$ This is exactly what you were asked to prove, since in a definite integral the name of the dummy variable of integration is irrelevant.

Comment: Since you doubted the reasonableness of the result, maybe the following geometric idea will be helpful. The integral $\int_1^x\frac{1}{t}\,dt$ is the area of the region $C$ which is below $y=\frac{1}{t}$, above the $t$-axis, from $t=1$ to $t=x$. Draw this region, maybe for a specific $x$ like $x=8$.

Now scale along the $t$-axis, by multiplying $t$-coordinates by $a$. At the same time, scale along the $y$-axis, by multiplying $y$-coordinates by $1/a$. If it will help, imagine that $a=3$. So we are stretching $C$ by the factor $3$ in the horizontal direction, and "stretching" by the factor $1/3$ (shrinking) in the vertical direction.

The double scaling turns a rectangle of base $u$ and height $v$ into a rectangle of base $au$ and height $v/a$. Notice that the area of the scaled rectangle is the same as the area of the original rectangle. So it is at least plausible (and even true) that our double scaling transforms the region $C$ into a region $C^\ast$ with the same area as $C$.

This new region $C^\ast$ goes from $t=a$ to $t=ax$, because of the stretching along the $t$ axis. And the height of $C^\ast$ at $t$ is $(1/a)$ times the height of $C$ at $t/a$. So the height of $C^\ast$ at $t$ is just $(1/a)(1/(t/a))$, which is simply $\frac{1}{t}$. Thus the area of $C^\ast$ is the area under $y=\frac{1}{t}$, from $t=a$ to $t=ax$. We have argued that the area of $C^\ast$ is the same as the area of $C$. That yields the desired integration result.

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If $\alpha$ is large, the integrand of $G$ is smaller than that of $F$ but the range of integration is much larger. In fact these two effects cancel, which is what you are asked to demonstrate. One approach, not the one in your problem is to substitute $u=\alpha t$ in $G$, getting $G(x) = \int_{a}^{ax}\frac{dt}{t}=\int_1^x \frac{du}{u}=F(x)$

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