Yes! The formula you've shown is a beautiful identity connecting integrals and discrete sums. It's an Euler–Maclaurin-like formula, sometimes used in numerical analysis and analytic number theory. Here's the identity again:

∫mnf(x) dx=∑k=0∞(−1)k(k+1)!∑j=m+1nf(k)(j)\int_m^n f(x)\, dx = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} \sum_{j=m+1}^{n} f^{(k)}(j)

This formula expresses a definite integral of a function ff in terms of the values of its derivatives at integer points. The proof uses Taylor expansions and the Euler–Maclaurin formula ideas. Let me walk you through a sketch of the proof.

Sketch of Proof

Let ff be a smooth function (say, C∞C^\infty) on the interval [m,n][m, n], and suppose we want to relate the integral ∫mnf(x) dx\int_m^n f(x) \, dx to sums over integer values.

1. Taylor Expansion around Integer Points

First, expand f(x)f(x) using a Taylor series about each integer j∈{m+1,m+2,…,n}j \in \{m+1, m+2, \dots, n\}:

f(x)=∑k=0∞f(k)(j)k!(x−j)kf(x) = \sum_{k=0}^\infty \frac{f^{(k)}(j)}{k!}(x - j)^k

Now integrate both sides over the interval [j−1,j][j-1, j]:

∫j−1jf(x) dx=∑k=0∞f(k)(j)k!∫j−1j(x−j)kdx\int_{j-1}^{j} f(x)\, dx = \sum_{k=0}^\infty \frac{f^{(k)}(j)}{k!} \int_{j-1}^{j} (x - j)^k dx

Let’s compute that inner integral:

∫j−1j(x−j)kdx=∫−10ukdu=(−1)kk+1\int_{j-1}^{j} (x - j)^k dx = \int_{-1}^{0} u^k du = \frac{(-1)^{k}}{k+1}

So:

∫j−1jf(x) dx=∑k=0∞(−1)k(k+1)!f(k)(j)\int_{j-1}^{j} f(x)\, dx = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} f^{(k)}(j)

2. Sum Over All Integer Intervals

Now, sum both sides over j=m+1j = m+1 to nn:

∫mnf(x) dx=∑j=m+1n∫j−1jf(x) dx=∑j=m+1n∑k=0∞(−1)k(k+1)!f(k)(j)\int_{m}^{n} f(x)\, dx = \sum_{j=m+1}^{n} \int_{j-1}^{j} f(x)\, dx = \sum_{j=m+1}^{n} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} f^{(k)}(j)

Now, change the order of summation (justified if ff is nice enough):

∫mnf(x) dx=∑k=0∞(−1)k(k+1)!∑j=m+1nf(k)(j)\int_{m}^{n} f(x)\, dx = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} \sum_{j=m+1}^{n} f^{(k)}(j)

And this is exactly the identity you showed.

Conclusion

✅ Proved:

∫mnf(x) dx=∑k=0∞(−1)k(k+1)!∑j=m+1nf(k)(j)\boxed{ \int_m^n f(x)\, dx = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} \sum_{j=m+1}^{n} f^{(k)}(j) }

This identity is a special case of series approximations of integrals. It’s also related to the Euler–Maclaurin formula but in a more symmetric form over integer points.

Let me know if you'd like a concrete example using a specific f(x)f(x) to illustrate this.