Can the series sum from n=1 to infinity frac1x²+n², x
Can the series $\displaystyle\sum_{n=1}^{\infty}\frac1{x^2+n^2}$, $x\in(-\infty,\infty)$, be integrated term by term?
Explanation of Method
Term-by-term integration on $[a,b]$ needs (1) each term continuous and (2) the series uniformly convergent. Both hold globally here: the series is dominated by $1/n^2$ (drop the non-negative $x^2$), which is summable, so the Weierstrass M-test gives uniform convergence on all of $\mathbb R$, and every term is a smooth rational function.
- Bound each term by $1/n^2$, uniform in $x$, and apply the Weierstrass M-test.
- Confirm each term is continuous, satisfying the second hypothesis.
- Integrate term by term and identify the resulting series.
Step 1 · Uniform convergence
Since $x^2\ge0$: $x^2+n^2\ge n^2$, so $\dfrac1{x^2+n^2}\le\dfrac1{n^2}$ for all $x\in\mathbb R$. Since $\sum1/n^2=\pi^2/6<\infty$, the Weierstrass M-test gives uniform (and absolute) convergence on $\mathbb R$.
Step 2 · Continuity
Each $u_n(x)=1/(x^2+n^2)$ is continuous on $\mathbb R$ (denominator never $0$), so it is continuous — hence integrable — on any interval $[a,b]$.
Step 3 · Term-by-term integration
Both hypotheses hold, so integration and summation may be swapped on any $[a,b]$: $$ \int_a^b\sum_{n=1}^{\infty}\frac1{x^2+n^2}\,dx = \sum_{n=1}^{\infty}\int_a^b\frac{dx}{x^2+n^2} = \sum_{n=1}^{\infty}\frac1n\left(\arctan\frac bn-\arctan\frac an\right), $$ using $\int dx/(x^2+n^2)=\frac1n\arctan(x/n)+C$. This integrated series converges, since for large $n$, $\arctan(b/n)-\arctan(a/n)\approx(b-a)/n$, giving a term of size $\sim(b-a)/n^2$.
Summary Table
| Condition | Status |
|---|---|
| Uniform convergence on $\mathbb R$ | Yes ($M_n=1/n^2$) |
| Continuity of each term | Yes |
| Term-by-term integration | Valid on any $[a,b]$ |
Additional Notes
- This is Problem 194's series revisited from the integration side: the same M-test bound $1/n^2$ that proved uniform convergence there now also licenses swapping the integral and the sum here.
- Unlike differentiation (Problems 216–218), where an extra decay factor was often needed to make the derivative series behave, integration is comparatively forgiving: it only ever needs the original series to converge uniformly, since $\int u_n\,dx$ is automatically at least as well-behaved as $u_n$ itself.
- The integrated series $\sum\frac1n\left(\arctan\frac bn-\arctan\frac an\right)$ is itself a telescoping-flavored construction reminiscent of Problem 213, expressing a definite integral as an infinite sum of arctangent differences.
Result
Yes — valid on any interval $[a,b]$: