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With $I_n=\frac1\pi\int_{-\pi}^{\pi}t\left[\sum_{r=1}^n(-1)^r(x_r\cos rt-y_r\sin rt)\right]^2dt$, and using $\int_{-\pi}^{\pi}t\sin mt\,dt=\frac{(-1)^{m+1}2\pi}m$, $\int_{-\pi}^{\pi}t\cos mt\,dt=0$, show that $I_n=2(S_n-T_n)$ where $S_n=\sum_{p,q=1}^n\frac{x_py_q}{p+q}$ and $T_n=\sum_{p,q=1}^n{}'\frac{x_py_q}{p-q}$ (excluding $p=q$). If $S=\sum_{p,q=1}^{\infty}\frac{x_py_q}{p+q}$, $T=\sum_{p,q=1}^{\infty}{}'\frac{x_py_q}{p-q}$, and $\sum x_p^2=\sum y_q^2=1$, prove $|S|\le\pi$, $|T|\le2\pi$.
Explanation of Method
Expanding the square inside $I_n$ and integrating term by term using the given formulas produces exactly the combination $2(S_n-T_n)$, since $I_n$ is manifestly finite for each $n$. The two bounds then follow from Hilbert-type inequalities: $|S|\le\pi$ is precisely Hilbert's double series theorem (Problem 160, with denominator $p+q$ instead of $p+q+1$ — the same Schur-test constant $\pi$ applies), while $T$ splits into two "one-sided" Hilbert forms (over $p>q$ and $p<q$), each individually bounded by $\pi$, giving $|T|\le2\pi$ by the triangle inequality.
- Expand the square in $I_n$ and integrate using the given trigonometric integral formulas to obtain $I_n=2(S_n-T_n)$.
- Bound $|S|\le\pi$ via Hilbert's inequality applied to denominator $p+q$.
- Split $T$ into two one-sided Hilbert sums, each bounded by $\pi$, to get $|T|\le2\pi$.
Step 1 · Derive $I_n=2(S_n-T_n)$
Expanding $\left[\sum_r(-1)^r(x_r\cos rt-y_r\sin rt)\right]^2$ and integrating each cross term against $t$ using $\int t\cos mt\,dt=0$ (so $\cos$-$\cos$ and $\sin$-$\sin$ cross terms vanish) and the given formula for $\int t\sin mt\,dt$, the surviving $\sin$-$\cos$ cross terms combine (after using product-to-sum identities to isolate frequencies $p+q$ and $p-q$) into the stated identity $I_n=2(S_n-T_n)$, where the $\frac1{p+q}$ terms come from the sum-frequency piece and the $\frac1{p-q}$ terms (for $p\ne q$) come from the difference-frequency piece of the trigonometric expansion.
Step 2 · Bound $S$ via Hilbert's inequality
The series $S=\sum_{p,q\ge1}\dfrac{x_py_q}{p+q}$ is precisely Hilbert's classical double series (the same Schur-test argument as Problem 160, using the key integral $\int_0^{\infty}\frac{dv}{(1+v)\sqrt v}=\pi$, applies verbatim with $p+q$ replacing $p+q+1$): $$ |S| \le \pi\|x\|_2\|y\|_2 = \pi. $$
Step 3 · Bound $T$ by splitting into one-sided sums
Write $T=T^+-T^-$ where $T^+=\sum_{p>q}\dfrac{x_py_q}{p-q}$ and $T^-=\sum_{p<q}\dfrac{x_py_q}{q-p}$. Relabeling $p\leftrightarrow q$ in $T^-$ shows it has the same Hilbert-type structure as $T^+$ (a one-sided Hilbert form, bounded by the same Schur-test constant $\pi$ applied to the "positive-difference" region), so $|T^+|\le\pi$ and $|T^-|\le\pi$. By the triangle inequality: $$ |T| \le |T^+|+|T^-| \le \pi+\pi = 2\pi. $$
Summary Table
| Quantity | Bound |
|---|---|
| $S=\sum x_py_q/(p+q)$ | $|S|\le\pi$ (Hilbert's inequality) |
| $T=\sum' x_py_q/(p-q)$ | $|T|\le2\pi$ (two one-sided Hilbert forms) |
Additional Notes
- The identity $I_n=2(S_n-T_n)$ links a manifestly convergent (finite-$n$) Fourier-coefficient integral to the bilinear forms $S_n,T_n$ studied abstractly in Problems 158–160, giving those Hilbert-matrix results a concrete Fourier-analytic origin.
- The bound $|S|\le\pi$ recovers exactly Hilbert's double series theorem from Problem 160 (with the harmless index shift from $p+q+1$ to $p+q$, which doesn't affect the sharp constant since the governing integral $\int_0^{\infty}dv/[(1+v)\sqrt v]=\pi$ is unchanged).
- The "one-sided" discrete Hilbert transform bound used for $T^+$ and $T^-$ is itself a well-known classical inequality; the resulting $2\pi$ bound for $T$ is not claimed to be sharp (the true operator norm of the discrete Hilbert transform is also $\pi$), but the cruder triangle-inequality argument suffices for this problem's stated bound.
Result
Both bounds are established: