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Find the following limit:

$$\large\lim_{n\to\infty} \int_{-\pi}^\pi \biggl(x + \frac{\pi}{2}\biggr)^2 \frac{\sin\bigl(\bigl(n+\frac{1}{2}\bigr)x\bigr) + x \cos nx}{\sin \frac{x}{2}}\,dx.$$

We tried to use what we know about the Dirichlet kernel and the real version of the Riemann-Lebesgue lemma but we're stuck.

Thanks

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  • $\begingroup$ Angle addition formula? $\endgroup$ – Mose Wintner Aug 6 '15 at 8:04
  • $\begingroup$ Split the fraction. For the one part, the Riemann Lebesgue lemma applies directly, for the other part, you have the Dirichlet kernel in it. $\endgroup$ – Daniel Fischer Aug 6 '15 at 8:13
  • $\begingroup$ We tried it but don't you need $f(x)$ to be 2π periodic to use the Riemann-Lebesgue lemma? $\endgroup$ – lfc Aug 6 '15 at 8:25
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The version of the Riemann-Lebesgue lemma we need here says that for an integrable function $f\colon [a,b] \to \mathbb{R}$, we have

$$\lim_{c\to \infty} \int_a^b f(x)\operatorname{trig}(cx)\,dx = 0,$$

where $\operatorname{trig}$ can be either of $\sin$ or $\cos$. It can be proved in many ways; we can extend the function trivially (setting $f(x) = 0$ for $x\notin [a,b]$) to $\mathbb{R}$ and appeal to the $L^1(\mathbb{R})$ version, or we can first show it for continuously differentiable $f$ by integration by parts, and then use that every integrable function can be approximated (in the integral sense) by continuously differentiable functions to obtain the result for all integrable functions on $[a,b]$.

Hence by the Riemann-Lebesgue lemma we have

$$\lim_{n\to\infty} \int_{-\pi}^\pi \biggl(x+\frac{\pi}{2}\biggr)^2 \frac{x}{\sin \frac{x}{2}}\cos nx\,dx = 0,$$

since $\bigl(x+\frac{\pi}{2}\bigr)^2\cdot\frac{x}{\sin (x/2)}$ is integrable on $[-\pi,\pi]$ - it is even continuously differentiable.

That leaves us to consider the part

$$\int_{-\pi}^\pi \biggl(x+\frac{\pi}{2}\biggr)^2\cdot \frac{\sin \bigl(\bigl(n+\frac{1}{2}\bigr)x\bigr)}{\sin \frac{x}{2}}\,dx.\tag{1}$$

The second factor in that is the Dirichlet kernel $D_n$. Now let us define

$$f(x) = \biggl(\frac{\pi}{2} - x\biggr)^2$$

for $x \in [-\pi,\pi)$, and extend that $2\pi$-periodically to $\mathbb{R}$. Then $f$ is a piecewise continuous and piecewise continuously differentiable function which is continuously differentiable on $\mathbb{R}\setminus (\pi + 2\pi\mathbb{Z})$. It has jump discontinuities at odd integer multiples of $\pi$. The general Fourier theory tells us that hence

$$f(t) = \lim_{n\to\infty} \frac{1}{2\pi} \int_{-\pi}^\pi f(t-x)D_n(x)\,dx\tag{2}$$

for all $t\in \mathbb{R}\setminus (\pi + 2\pi\mathbb{Z})$. But the integral in $(1)$ is just

$$\int_{-\pi}^\pi f(0-x)D_n(x)\,dx,$$

so by $(2)$ we have

$$\lim_{n\to\infty} \int_{-\pi}^\pi \biggl(x+\frac{\pi}{2}\biggr)^2\cdot \frac{\sin \bigl(\bigl(n+\frac{1}{2}\bigr)x\bigr)}{\sin \frac{x}{2}}\,dx = 2\pi\cdot f(0) = 2\pi\biggl(\frac{\pi}{2}-0\biggr)^2 = \frac{\pi^3}{2}.$$

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