Cauchy's formula for repeated integration tells us that the $n$'th repeated integral of a function $f$, evaluated from $a$ to $x$, can be expressed as a single integral:
$$f^{(-n)}(x)=\frac{1}{(n-1)!}\int_{a}^{x} (x-t)^{n-1}f(t)dt$$
Using the exponent function as an example:
$$\int_{-\infty}^{x}e^{t}dt = e^{x}$$
How would one go about deriving $f(x)=e^{x}$, given its identity equation:
$$f(x)=\frac{1}{(n-1)!}\int_{-\infty}^{x} (x-t)^{n-1}f(t)dt$$
In other words, how can one isolate and solve for $f$ in the following equation:
$$e^{x}=\frac{1}{(n-1)!}\int_{-\infty}^{x} (x-t)^{n-1}f(t)dt$$
rather than plugging in $f(t)=e^{t}$, and showing that it is a valid solution?
I have tried the following simple steps, and I don't know how to proceed:
- Differentiate both sides with respect to $x$
$$\frac{\text{d}}{\text{d}x}(e^{x})=\frac{\text{d}}{\text{d}x}\left(\frac{1}{(n-1)!}\int_{-\infty}^{x} (x-t)^{n-1}f(t)dt\right)$$
2)
$$e^{x}=\lim_{a \to -\infty}\frac{1}{(n-1)!}\left.(x-t)^{n-1}f(t)\right|_{t=a}^{t=x}$$
- We now lose $f(x)$ as it is being multiplied by $(x-x)^{n-1}$ i.e., $0$
$$e^{x}=\lim_{a \to -\infty}\frac{1}{(n-1)!}\left((x-x)^{n-1}f(x) - (x-a)^{n-1}f(a)\right)$$
4)
$$\lim_{a \to -\infty}f(a)\frac{-(x-a)^{n-1}}{(n-1)!} = e^{x}$$
This equation only seems to give us an expression for one point $f(a)$ as $a\rightarrow-\infty $
To summarise, can we somehow find an expression for $f(x)$ from $e^{x}=\frac{1}{(n-1)!}\int_{-\infty}^{x} (x-t)^{n-1}f(t)dt$
Thank you!