Show $\int_0^t f(s)(\int_0^sf(m)dm )^k ds = \frac{1}{k+1}( \int_0^tf(s)ds)^{k+1}$

Question

Is true that$\displaystyle\int_0^t f(s)\left(\displaystyle\int_0^sf(m)dm \right)^k ds = \dfrac{1}{k+1}\left( \displaystyle\int_0^tf(s)ds\right)^{k+1}$ for $f$ continuous positive?

I've seen this equality being used in an PDE book, but deriving the right side and using the fundamental theorem of calculus, it doesn't seem to do the same thing. I'm wrong?

Answer 1

Taking the derivative of both sides with respect to $t$ gives

$$\frac{d}{dt}\int_0^t f(s)\left(\displaystyle\int_0^sf(m)dm \right)^k ds = f(t)\left(\displaystyle\int_0^tf(m)dm \right)^k $$

$$\frac{d}{dt}\dfrac{1}{k+1}\left( \displaystyle\int_0^tf(s)ds\right)^{k+1} = \left( \displaystyle\int_0^tf(s)ds\right)^{k}f(t)$$

which are exactly the same (the integration variable is just a dummy variable). This means that the original expressions are equivalent, except possibly off by a constant. In order to see what that constant is, plug in $t=0$ on both sides:

$$\int_0^0 f(s)\left(\displaystyle\int_0^sf(m)dm \right)^k ds = 0$$

$$\dfrac{1}{k+1}\left( \displaystyle\int_0^0 f(s)ds\right)^{k+1} = 0$$

so the difference between the two values at $t=0$ is $0$, meaning the expressions are exactly equivalent.

Answer 2

Let $$F(s) = \int_0^s f(m)dm$$ Then, since $f(s)$ is continuous, we have $dF(s) = F'(s)ds = f(s)ds$. Thus, we can rewrite the LHS as $$\begin{align} \int_0^t F^k(s)dF(s) &= \frac{1}{k+1}F^{k+1}(s)\bigg|_0^t \\ &= \frac{1}{k+1}[F^{k+1}(t) - \underbrace{F^{k+1}(0)}_0] \\ &= \frac{1}{k+1}F^{k+1}(t)\end{align}$$

---

By induction, using the same approach, we can generalize this as $$\int_0^t dt_1\int_{0}^{t_1}dt_2\int_0^{t_2}dt_3\dots\int_{0}^{t_{n-1}} f(t_1)f(t_2)\dots f(t_n)dt_n = \frac{1}{n!}\left(\int_0^t f(\tau)d\tau\right)^n$$