Inverse of a Function Under an Integral

Question

If a function is defined as the integral of another function, then is there a simple procedure to find the inverse of that function?

For Example;

$f(x)= \int_2^x ln(t) dt$

From there can we use that inverse to solve for the value of $(f^{-1})'(0)$? Also, does it make a difference if the integral is definite or indefinite?

Answer 1

This answer assumes knowledge of the following:

• Fundamental Theorem of Calculus (FTC). In particular, we are looking at the second fundamental theorem of calculus, which states that: If $f$ is a continuous function on an open interval $I$ and $a$ is any point in $I$, then for $F$ defined as: $$F(x)=\int_{a}^{x}f(t) \ dt$$

We will have that $F'(x)=f(x)$ for each point in $I$.

• The inverse of any function $f(x)$ exists only if $f(x)$ is injective, that is, $f(x_1)=f(x_2) \Rightarrow x_1=x_2 \forall \ x_1,x_2 \in \mathbb{R}$. If the inverse of $f(x)$ exists, we will denote it as $f^{-1}(x)$. In addition, we will denote the domain and range of $f(x)$ as $D_f$ and $R_f$ respectively, and we will denote the domain and range of $f^{-1}(x)$ as $D_{f^{-1}}$ and $R_{f^{-1}}$ respectively. Then, we have the following: $$D_f=R_{f^{-1}}$$ and $$R_f=D_{f^{-1}}$$ In other words, the domain of the inverse function is precisely the range of $f(x)$, while the range of $f^{-1}(x)$ is the domain of $f(x)$.
• Understanding the idea of function composition.
• Chain Rule for differentiation

Armed with these knowledge, we can start tackling the problem (I am referring to the part about evaluating the value of $(f^{-1})'(0)$, and not the more general problem that you posed). In your question, it is given that $f(x)=\int_{a}^{x}ln(t) \ dt $. Clearly, $D_f=[0, \infty]$. Now, we need to check whether the function is injective over $[0,\infty]$; otherwise, the inverse might not exist over the domain $D_f$, and it becomes meaningless to compute the value of $(f^{-1})'(0)$.

Let $g(t)=ln(t)$, such that $f(x)=\int_{a}^{x} g(t) \ dt$. Applying the Fundamental Theorem of Calculus, we have: $f'(x)=g(x)=ln(x)$. So here's how to evaluate the derivative of $f(x)$ without first having to integrate $ln(x)$ by parts and then differentiating the result!

Clearly, $f(x)$ has only one stationary point, which occurs when $ln(x)=0$, i.e. $x=1$. Thus, $f(x)$ is not injective over $[0, \infty]$, which means the inverse is not defined for this domain.

Yet, $f^{-1}(x)$ can exist as long as we restrict the domain of $f(x)$. In particular, we if we restrict the domain of $f(x)$ to $[0,1]$ or $[1,\infty]$ such that it becomes injective over the new restricted domain, then $f^{-1}$ is defined. So, we can either have $D_f=[0,1]$ or $D_f=[1, \infty]$.

Case 1

$D_f=[0,1]$. We have that $f^{-1}(f(x))=x, \ \forall \ x \in [0,1]$. To find the value of $f^{-1}(0)$, simply set $f(x)=0$. Plotting the graph of $f(x)$ out, we can see that it has $2$ roots. By inspection, $x=2$ will be a root (Can you see why?). Unfortunately, the other root cannot be solved analytically , so we use Wolfram Alpha to obtain $ x \approx 0.263$ (I don't intend to invoke the symbolic use of the Lambert-W function here). But we have to reject $x=2$ in this case, simply because it does not lie in the interval $[0,1]$. Hence, we must have that $f^{-1}(f(0.263))=0.263=f^{-1}(0)$.

In addition, $f(f^{-1}(x))=x \ \forall \ x \in D_{f^{-1}}=R_f$. Applying the chain rule, we have that $f'(f^{-1}(x)) \cdot (f^{-1})'(x)=1$. Thus, $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \Rightarrow (f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}=\frac{1}{f’(0.263)}=\frac{1}{ln(0.263)}$.

Case 2

$D_f=[1, \infty]$. Now, we take the root $x=2$ and reject $x \approx 0.263$ instead. Hence,$f^{-1}(f(2))=2=f^{-1}(0)$.

In addition, $f(f^{-1}(x))=x \ \forall \ x \in D_{f^{-1}}=R_f$. Applying the chain rule, we have that $f'(f^{-1}(x)) \cdot (f^{-1})'(x)=1$. Thus, $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \Rightarrow (f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}=\frac{1}{f’(2)}=\frac{1}{ln(2)}$.

Note that these are not the only unique ways to restrict the domain of $f(x)$ such that it is injective. The purpose of me doing so was to illustrate how to find the $2$ possible values for $(f^{-1})'(x)$.

I hope this answers your question, at least in part.