In Casella and Berger Book (Statistical Inference), exercise 2.40 is
$$\sum_{k=0}^x {n\choose k}p^k(1-p)^{n-k}=(n-x){n\choose x}\int_0^{1-p}t^{n-x-1}(1-t)^xdt.$$
If I replace $x$ by $n$ then LHS becomes $1$ but RHS becomes $0$. Am I missing something?
On
Note that we have
$$\begin{align} \lim_{x\to n^-}\sum_{k=0}^x p^k(1-p)^{n-k}&=\lim_{x\to n^-}(n-x)\binom{n}{x}\int_0^{1-p}t^{n-x-1}(1-t)^x\,dt\\\\ &=\lim_{y\to 0^+}\left(y\frac{\Gamma(n+1)}{\Gamma(n+1-y)\Gamma(y+1)}\int_0^{1-p}t^{y-1}(1-t)^{n-y}\,dt\right) \tag 1 \end{align}$$
Now, let's examine the limit,
$$L=\lim_{y\to 0^+}\left(y\int_0^{1-p}t^{y-1}(1-t)^{n-y}\,dt\right) \tag 2$$
Integrating by parts the integral in $(2)$ with $u=(1-t)^{n-y}$ and $v=\frac{t^y}{y}$ we find
$$\begin{align} L&=\lim_{y\to 0^+}\left(y\int_0^{1-p}t^{y-1}(1-t)^{n-y}\,dt\right)\\\\ &=\lim_{y\to 0^+}\left(p^{n-y}(1-p)^y+(n-y)\int_0^{1-p}t^y(1-t)^{n-y-1}\,dt\right)\\\\ &=p^n+n\int_0^{1-p}(1-t)^{n-1}\,dt\\\\ &=1 \tag 3 \end{align}$$
Inasmuch as $\lim_{y\to 0^+}\frac{\Gamma(n+1)}{\Gamma(n+1-y)\Gamma(y+1)}=1$, then using $(3)$ in $(2)$ yields
$$\lim_{x\to n^-}(n-x)\binom{n}{x}\int_0^{1-p}t^{n-x-1}(1-t)^x\,dt=1$$
Therefore, the expression of interest is correct when removing the discontinuity at $n=x$ by using the limiting value of $1$.
My instinct is that the RHS will actually be $0 \cdot \infty$ since your integral will no longer converge. So, rather than being an incorrect equation, it'll just become an ill-posed one. :D