Proof of an inverse If $F(x)$ = $\int\limits_1^x$ $\frac{1}{t}$ $dt$ then the function $F$ has an inverse.
I know that $F(x)$ = $\int\limits_1^x \frac{1}{t} = ln x$ but I can't assume that it is $lnx$. So I really don't know how to begin this proof and direction on it would be nice.
 A: We need to show that $F$ is injective and surjective.
Injectivity is easy, as $F$ is the integral of a positive function and hence increasing: If x' > x, then
$$F(x') = \int_1^{x'} \frac{dt}{t} = \int_1^x \frac{dt}{t} + \int_x^{x'} \frac{dt}{t} > \int_0^x \frac{dt}{t} = F(x).$$
Surjectivity takes a little more work: For any integer $n > 0$, the right-hand estimate gives
$$\int_1^n \frac{dt}{t} \geq \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n},$$ but the sum on the right diverges as $n \to \infty$.
So, $F$ takes on arbitrarily large positive values, and since, e.g., $F(1) = 0$, by the Intermediate Value Theorem it takes on all nonnegative values.
To show that $F$ takes on all negative values, observe the following: Substituting $t = \frac{1}{u}$, $dt = -\frac{du}{u^2}$ gives
$$\int_1^{1/x} \frac{1}{\left(\frac{1}{u}\right)} \left(-\frac{du}{u^2}\right) = - \int_1^{1/x} \frac{du}{u}.$$ So, by the argument for the positive case, this integral (without the negative sign) takes on the values $[0, \infty)$ for $1 \leq \frac{1}{x} < \infty$, that is, for $0 < x \leq 1$, and so the quantity (now including the negative sign) takes on the values $(-\infty, 0]$ as $x$ varies over that set. The two cases together gives that $F$ is surjective.
