Find integral from 0 to infinity (sum from n=1
Find $\displaystyle\int_0^{\infty}\left(\sum_{n=1}^{\infty}\frac1{n^4+x^2}\right)dx$.
Explanation of Method
Since every term is non-negative, Tonelli's theorem licenses interchanging the sum and the (improper) integral without any separate uniform-convergence check: computing $\sum_n\int_0^\infty u_n(x)\,dx$ directly is itself the proof that the interchange is valid, since a finite value for that sum automatically equals the double integral in either order.
- Compute the individual integral $\int_0^{\infty}\frac{dx}{n^4+x^2}$ via substitution.
- Sum the resulting series over $n$.
- Justify the interchange via Tonelli's theorem for non-negative integrands.
Step 1 · Individual integrals
Substitute $u=x/n^2$, $dx=n^2\,du$: $$ \int_0^{\infty}\frac{dx}{n^4+x^2} = \int_0^{\infty}\frac{n^2\,du}{n^4(1+u^2)} = \frac1{n^2}\int_0^{\infty}\frac{du}{1+u^2} = \frac1{n^2}\left[\arctan u\right]_0^{\infty} = \frac{\pi}{2n^2}. $$
Step 2 · Sum over $n$
$$ \sum_{n=1}^{\infty}\frac{\pi}{2n^2} = \frac{\pi}2\sum_{n=1}^{\infty}\frac1{n^2} = \frac{\pi}2\cdot\frac{\pi^2}6 = \frac{\pi^3}{12}. $$
Step 3 · Justifying the interchange
Every term $u_n(x)=1/(n^4+x^2)$ is non-negative on $[0,\infty)$. By Tonelli's theorem (the non-negative-function form of Fubini's theorem), summing and integrating a non-negative double array can always be interchanged, and the resulting value is automatically the same in either order — no uniform-convergence hypothesis is needed. Since Step 2 shows $\sum_n\int_0^\infty u_n\,dx=\pi^3/12<\infty$, this value equals $\int_0^\infty\sum_n u_n(x)\,dx$ directly.
Summary Table
| Quantity | Value |
|---|---|
| $\int_0^{\infty}dx/(n^4+x^2)$ | $\pi/(2n^2)$ |
| Final answer | $\pi^3/12$ |
Additional Notes
- Tonelli's theorem is the natural tool here because every term is non-negative — unlike Fubini's theorem, it requires no separate absolute-integrability check; a finite value for the iterated sum-then-integrate automatically certifies the interchange is valid.
- This series $\sum1/(n^4+x^2)$ is a special case ($m=4$) of the general family $\sum1/(n^m+x^2)$ studied throughout this problem set (Problems 194 and 208 use $m=3,4$ respectively for finite-sum bounds); here the same family is integrated over all of $x\ge0$ rather than evaluated pointwise.
- The final answer $\pi^3/12$ combines the arctangent integral $\int_0^\infty dx/(1+u^2)=\pi/2$ with the Basel sum $\sum1/n^2=\pi^2/6$ — two of the most recurring constants across this entire problem collection.