Leibniz Integral Rule applied n-times

Question

I was trying to come up with a generalization of the Leibniz Integral Rule $$ \frac{d}{dx}\left(\int_a^xf(x,t) \; dt\right) = f(x,x) + \int_a^x \frac{\partial}{\partial x} f(x,t) \; dt $$ and arrived at the following and i am not sure if it is correct $$ \frac{d^n}{dx^n}\left(\int_a^xf(x,t) \; dt\right) = \sum_{k=0}^{n-1}f^{(k)}(x,x)+\int_a^x \frac{\partial^n}{\partial x^n} f(x,t) \; dt $$ Can someone maybe confirm or refute this?

Answer 1

After computing the first few derivatives you can immediately see, that the pattern is the following: $$\frac{d^n}{dx^n}\left(\int_a^xf(x,t) dt\right) = \sum_{k=0}^{n-1}\binom{n}{k+1}\frac{\partial^{n-1} f}{\partial x^{n-1-k}\partial t^k}(x,x) + \int_a^x\frac{\partial ^n}{\partial x^n}f(x,t)dt$$ Trivially in the case $n=1$ this reduces to the known formula. Now for the induction step: \begin{align*} \frac{d^{n+1}}{dx^{n+1}}\left(\int_a^xf(x,t) dt\right) =& \frac{d}{dx}\left(\sum_{k=0}^{n-1}\binom{n}{k+1}\frac{\partial^{n-1} f}{\partial x^{n-1-k}\partial t^k}(x,x) + \int_a^x\frac{\partial ^n}{\partial x^n}f(x,t)dt\right)\\ =& \sum_{k=0}^{n-1}\binom{n}{k+1}\frac{d}{dx}\frac{\partial^{n-1} f}{\partial x^{n-1-k}\partial t^k}(x,x) + \frac{d}{dx}\int_a^x\frac{\partial ^n}{\partial x^n}f(x,t)dt\\ =& \sum_{k=0}^{n-1}\binom{n}{k+1}\left(\frac{\partial^{n} f}{\partial x^{n-k}\partial t^k}(x,x)+\frac{\partial^{n} f}{\partial x^{n-1-k}\partial t^{k+1}}(x,x)\right)\\ & + \frac{\partial^{n} f}{\partial x^{n}}(x,x)+\int_a^x\frac{\partial ^{n+1}}{\partial x^{n+1}}f(x,t)dt \end{align*} So it remains to show that \begin{align*} \sum_{k=0}^{n-1}\binom{n}{k+1}\left(\frac{\partial^{n} f}{\partial x^{n-k}\partial t^k}(x,x)+\frac{\partial^{n} f}{\partial x^{n-1-k}\partial t^{k+1}}(x,x)\right) + \frac{\partial^{n} f}{\partial x^{n}}(x,x) = \sum_{k=0}^{n}\binom{n+1}{k+1}\frac{\partial^{n} f}{\partial x^{n-k}\partial t^k}(x,x) \end{align*} For shorter notation, we set $$c_{n,k}:=\frac{\partial^{n} f}{\partial x^{n-k}\partial t^k}$$ so that above equation becomes \begin{align*} \sum_{k=0}^{n-1}\binom{n}{k+1}\left(c_{n,k}+c_{n,k+1}\right) + c_{n,0} = \sum_{k=0}^{n}\binom{n+1}{k+1}c_{n,k} \end{align*} This is now pure combinatorics: \begin{align*} \sum_{k=0}^{n-1}\binom{n}{k+1}\left(c_{n,k}+c_{n,k+1}\right) + c_{n,0} =& \sum_{k=0}^{n-1}\binom{n}{k+1}c_{n,k}+\sum_{k=0}^{n-1}\binom{n}{k+1}c_{n,k+1} + c_{n,0}\\ =& \sum_{k=0}^{n-1}\binom{n}{k+1}c_{n,k}+\sum_{k=1}^{n}\binom{n}{k}c_{n,k} + c_{n,0}\\ =& \sum_{k=0}^{n-1}\binom{n}{k+1}c_{n,k}+\sum_{k=0}^{n}\binom{n}{k}c_{n,k}\\ =& \sum_{k=0}^{n-1}\left(\binom{n}{k+1}+\binom{n}{k}\right)c_{n,k}+c_{n,n}\\ =&\sum_{k=0}^{n-1}\binom{n+1}{k+1}c_{n,k}+c_{n,n}\\ =&\sum_{k=0}^{n}\binom{n+1}{k+1}c_{n,k}. \end{align*} Thus by induction the above identity holds.

Answer 2

Here is another derivation of the formula obtained in @StiftungWarentest's answer.

For the sake of simplicity, assume that $f(x, t)$ is smooth and compactly supported on $\mathbb{R}^2$. Then by writing $\theta(t) = \mathbf{1}_{[0,\infty)}(t)$ for the Heaviside step function, the $n$th weak derivative of the integral is computed as

\begin{align*} \frac{\mathrm{d}^n}{\mathrm{d} x^n} \left[ \int_{a}^{x} f(x, t) \, \mathrm{d}t \right] &= \frac{\mathrm{d}^n}{\mathrm{d} x^n} \left[ \int_{a}^{\infty} f(x, t) \theta(x-t) \, \mathrm{d}t \right] \\ &= \int_{a}^{\infty} \frac{\partial^n}{\partial x^n} [ f(x, t) \theta(x-t) ] \, \mathrm{d}t \tag{1} \\ &= \sum_{k=0}^{n} \binom{n}{k} \int_{a}^{\infty} \left[\frac{\partial^{n-k} f}{\partial x^{n-k}} (x, t) \right] \theta^{(k)} (x-t) \, \mathrm{d}t. \tag{2} \end{align*}

• In $\text{(1)}$, we utilized the Leibniz integral rule.
• In $\text{(2)}$, we invoked the Leibniz product rule.

On the other hand, for any $n \geq 0$ and for any compactly supported smooth $\varphi$ on $\mathbb{R}$, a similar computation shows that

$$ \int_{a}^{\infty} \varphi(t)\theta^{(n)}(x-t) \, \mathrm{d}t = \begin{cases} \varphi^{(n-1)}(x), & n \geq 1, \\[0.5em] \displaystyle \int_{a}^{x} \varphi(t) \, \mathrm{d}t, & n = 0. \end{cases} $$

So by fixing $x$ and applying this identity to $\text{(2)}$, we obtain

\begin{align*} \frac{\mathrm{d}^n}{\mathrm{d} x^n} \left[ \int_{a}^{x} f(x, t) \, \mathrm{d}t \right] &= \sum_{k=1}^{n} \binom{n}{k} \frac{\partial^{n-1} f}{\partial x^{n-k}\partial t^{k-1}} (x, x) + \int_{a}^{x} \frac{\partial^{n} f}{\partial x^{n}} (x, t) \, \mathrm{d}t. \end{align*}

Finally, although the above equality holds a priori in weak derivative sense, both sides are bona-fide smooth functions for any $n$. Consequently, the equality holds in the usual derivative sense, yielding the correct formula for the $n$th derivative of the integral.