When Do Differentiation and Integration Undo Each Other?
A derivative removes constant terms, while an indefinite integral introduces a choice of constant. For power series, these familiar facts have a precise coefficient-level description: differentiation shifts and rescales coefficients, and integration shifts them back while leaving one coefficient to be chosen.
Keeping track of that coefficient matters whenever you use one operation to reverse the other. Differentiating an antiderivative recovers the original series exactly. Integrating a derivative recovers the original series only after its constant term is restored. Repeating the operations makes the same issue visible for a whole polynomial of low-degree terms.
Coefficient Rules for One Operation
Let a power series centered at \(a\) be
and suppose its radius of convergence is \(R>0\). The termwise differentiation and integration theorems apply for \(|x-a|<R\). Within that interval, differentiation gives
The second expression reindexes the differentiated series so that it is again written in powers \((x-a)^n\). In particular, the coefficient of \((x-a)^n\) after differentiation is \((n+1)c_{n+1}\), and the original constant coefficient \(c_0\) does not appear.
An antiderivative is obtained by integrating each term and adding an arbitrary constant \(C\):
Here the constant term of \(G\) is \(C\), while the coefficient of \((x-a)^{n+1}\) is \(c_n/(n+1)\). This integrated series has the same radius of convergence \(R\); the formulas are being used on its interior, \(|x-a|<R\).
Why the Two Orders Give Different Results
Let \(D\) denote termwise differentiation and let \(I_C\) denote termwise integration with the constant chosen to be \(C\). Applying differentiation to an antiderivative cancels the factor introduced by integration. Applying integration to a derivative restores every positive-degree coefficient, but it cannot recover the original constant coefficient unless that value is selected.
These identities hold for \(|x-a|<R\). In the second identity, \(c_0=F(a)\), since all positive powers of \(x-a\) vanish at \(x=a\). Thus choosing \(C=c_0\) makes the second composition recover \(F\) exactly; choosing \(C=0\) produces \(F-c_0\).
Why These Identities Hold
The identities follow by comparing coefficients, and the verification also explains why the constant term is the only term lost in a single differentiation.
The coefficient of \((x-a)^n\) in \(D F\) is \((n+1)c_{n+1}\). The coefficient \(c_0\) contributes no term because the derivative of a constant is zero.
Integrating \((n+1)c_{n+1}(x-a)^n\) gives \(c_{n+1}(x-a)^{n+1}\), since \((n+1)/(n+1)=1\). Therefore all original coefficients \(c_1,c_2,\ldots\) return in their original degrees. The integration constant is \(C\), so the result is \(C+\sum_{n=1}^{\infty}c_n(x-a)^n=F-c_0+C\).
Integrating \(F\) gives \(C+\sum_{n=0}^{\infty}c_n(x-a)^{n+1}/(n+1)\). Differentiating a term in this sum multiplies its coefficient by \(n+1\) and reduces its degree by one. Each term therefore becomes \(c_n(x-a)^n\), while the constant \(C\) differentiates to zero. The result is exactly \(F\).
The termwise operations are valid inside the original radius of convergence, where the power-series differentiation and integration theorems apply. No endpoint conclusion is needed for these identities.
Worked Example: A Polynomial in Both Orders
Worked Example: A Polynomial in Both Orders
Let \(t=x-2\), and consider \(F(x)=4-3t+5t^2-2t^3\). Differentiate term by term:
Integrate this derivative, keeping the arbitrary constant visible:
To recover \(F\), choose \(C=4\), the original constant coefficient. The result is \(4-3t+5t^2-2t^3=F(x)\). If instead \(C=0\), the result is \(F(x)-4\).
Check the reverse order by integrating \(F\):
Differentiate to check:
because \(dt/dx=1\). The arbitrary constant disappears, and every term of \(F\) returns.
Worked Example: An Infinite Series
Worked Example: An Infinite Series
Consider the power series \(F(x)=\sum_{n=0}^{\infty}x^n/n!\), which converges for every real \(x\). Differentiate term by term and reindex using \(m=n-1\):
Now integrate this derivative term by term, with integration constant \(C\):
Reindex the final sum with \(m=n+1\). It contains all terms of \(F\) except the degree-zero term:
Thus \(C=1\) recovers \(F\), whereas \(C=0\) gives \(F-1\). To check the opposite order, integrate \(F\) first:
Its derivative is \(\sum_{n=0}^{\infty}x^n/n!=F(x)\), as required. The coefficient calculation checks both compositions, including the constant term.
Repeated Operations and Lost Low-Degree Terms
The same principle extends to repeated operations. Differentiating \(k\) times removes all terms of degree less than \(k\), because each such term becomes zero before the \(k\)th derivative. Integrating \(k\) times introduces an arbitrary polynomial of degree at most \(k-1\): each integration contributes a new constant, and later integrations can turn those constants into higher-degree terms.
For \(F(x)=\sum_{n=0}^{\infty}c_n(x-a)^n\), the \(k\)th derivative, for a positive integer \(k\), is
If this series is integrated \(k\) times using zero constants at each stage, the result is \(F(x)-\sum_{j=0}^{k-1}c_j(x-a)^j\). The terms of degree \(k\) and above return, but the first \(k\) coefficients do not. Choosing the integration constants can restore those missing terms.
Worked Example: Two Derivatives and Two Integrations
Let \(t=x-1\) and \(F(x)=2+3t-4t^2+5t^3+t^4\). Differentiate twice:
Integrate \(F''\) once, choosing zero constant:
Integrate again, again choosing zero constant:
This is \(F(x)-2-3t\): the constant and linear terms were lost. Add the polynomial \(2+3t\) to recover \(F\). Check by differentiating the recovered expression twice:
The check confirms that two derivatives retain the terms of degree two and higher, while the two integration constants allow the missing degree-zero and degree-one terms to be chosen.
Common Pitfall: Assuming Integration Automatically Restores the Constant
A frequent error is to integrate a derivative and claim that the original function has been recovered without accounting for its constant term. For the polynomial in the first example, setting \(t=x-2\) gives
An incorrect calculation then says
The final equality is false: the expression on the left has constant coefficient \(0\) in powers of \(t\), while the expression on the right has constant coefficient \(4\). The correction is to retain the arbitrary constant:
Only when \(C=4\) does this equal the original polynomial. The general rule is not that integration reverses differentiation without qualification; it is that integration recovers the differentiated series up to an additive constant.
Why the Coefficient View Is Useful
These rules make it possible to check power-series calculations without relying only on a remembered function formula. Track the degree and coefficient of each term: differentiation lowers the degree by one and multiplies by the former degree; integration raises the degree by one and divides by the new degree. Then account for the constant that differentiation removes or integration leaves undetermined.
For repeated operations, identify how many low-degree coefficients have been removed. After \(k\) derivatives, terms of degrees \(0\) through \(k-1\) are absent. Integrating \(k\) times can restore them only if the resulting polynomial of degree at most \(k-1\) is selected. This distinction is the coefficient-level counterpart of the fact that a derivative cannot distinguish between functions that differ by a constant, or, after repeated differentiation, by a polynomial of the corresponding lower degree.
Check Your Understanding
Use coefficient tracking to answer each question. In questions involving an antiderivative, state what choice of constant or polynomial is needed for exact recovery.
- If \(F(x)=\sum_{n=0}^{\infty}c_n(x-a)^n\), what is the coefficient of \((x-a)^n\) in \(F'(x)\)?
- What does integrating \(F'\) with constant \(C=0\) give in terms of \(F\) and \(c_0\)?
- Why does differentiating an antiderivative of \(F\) recover \(F\) for any choice of integration constant?
- After taking two derivatives of a power series, which original degrees have disappeared?
- For \(F(x)=7+2(x-a)+(x-a)^2\), what polynomial must be added after integrating \(F''\) twice with zero constants to recover \(F\)?
