Why is this integral zero?

Question

I was solving a problem on Fourier Analysis ( namely, I was proving that $\int|f|^2dx=\sum_{n\in \mathbb{Z}} |f(n)|^2$ if $f$ is of moderate decrease and the fourier transform of $f$ is supported on $[-1/2,1/2]$) and the problem boils down to proving that this integral is zero for $n\not=m$ where $n$ and $m$ are both integers:

$$\int_{-\infty}^\infty \frac{\sin(\pi x)}{(x-n)}\frac{\sin(\pi x)}{(x-m)}dx=0$$

I am not really sure how to compute this integral. I know $\sin(\pi x)^2/x^2$ can be integrated using the fourier transform of a characteristic function. I tried to adapt this here but to no avail.

Answer 1

The function $f_{k}(x)=\mathbf{1}_{\left [-{1\over 2},{1\over 2}\right ]}e^{2\pi i kx}$ is square integrable and its Fourier transform is equal $$\widehat{f_k}(t)={1\over \pi (k-t)}\sin\pi (k-t)=(-1)^k{\sin\pi t\over \pi(t-k)}$$ By the Plancherel identity $$\int\limits_{-\infty}^\infty f_n(x)\overline{f_m(x)}\,dx =\int\limits_{-\infty}^\infty \widehat{f_n}(t)\overline{\widehat{f_m}(t)}\,dt\\ ={(-1)^{n+m}\over \pi^2}\int\limits_{-\infty}^\infty{\sin\pi t\over t-m}{\sin \pi t\over t-n}\,dt$$ The left hand side integral vanishes for $n\neq m,$ as the functions $ e^{2\pi i mx}$ and $ e^{2\pi i nx}$ are orthogonal in $L^2\left (-{1\over 2},{1\over 2}\right )$ for $m\neq n.$ This gives the conclusion.

Answer 2

I used $f(x)=\sum_{n\in \mathbb{Z}} {f}(n)\frac{\sin(\pi(x-n))}{\pi(x-n)}$ and expanded $|f(x)|^2$ using the fact this series is commutatively convergent to yield the expression. But this results in some cumbersome integrals and it is a bit tedious to prove our relation this way. Of course this initial approach also yields the right answer, but it ocurred to me there is a more natural way to do this!

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By Plancharel and also using that $\hat{f}$ has compact support in $[-1/2,1/2]$:

$$\int_{-\infty}^\infty |f(x)|^2dx=\int_{-\infty}^\infty |\hat{f}(w)|^2dw=\int_{-1/2}^{1/2}|\hat{f}(w)|^2$$

We may view the restriction of $\hat{f}$ as a continuous function on the circle. In this case, we have that is integrable in the circle and by Parseval we have:

$$\int_{-1/2}^{1/2}|\hat{f}(w)|^2dw=\sum_{n\in \mathbb{Z}}|a_n|^2 $$

Where $a_n$ is given by:

$$a_n=\int_{-1/2}^{1/2}\hat{f}(w)e^{-2\pi i n w}dw=\int_{-\infty}^\infty \hat{f}(w)e^{2\pi i (-n) w}dw={f}(-n)$$

Here we have used the Inversion Formula ($f$ and $\hat{f}$ are of moderate decrease). Substituting this expression in the equalities above yields:

$$\int_{-\infty}^\infty |f(x)|^2dx=\sum_{n\in \mathbb{Z}}|{f}(n)|^2$$

Answer 3

We can use partial fractions to break up the denominator: $$ \frac{1}{\left(x-n\right)\left(x-m\right)} = \frac{1}{n-m}\left(\frac{1}{x-n}-\frac{1}{x-m}\right). $$

Therefore, \begin{align} \int_{-\infty}^\infty \frac{\sin(\pi x)}{(x-n)}\frac{\sin(\pi x)}{(x-m)}dx &= \frac1{n-m} \int_{-\infty}^\infty \left(\frac{\sin^2(\pi x)}{x - n} - \frac{\sin^2(\pi x)}{x - m}\right)dx \\ &= \lim_{a,b \to \infty} \frac1{n-m} \left(\int_{-a}^b \frac{\sin^2(\pi x)}{x - n}dx - \int_{-a}^b\frac{\sin^2(\pi x)}{x - m}dx\right). \end{align} For the first integral, let $u = x + n$, and for the second, let $u = x + m$. Noting that in both cases $sin^2(\pi x) = sin^2(\pi u)$, we get $$ \lim_{a,b \to \infty} \frac1{n-m} \left(\int_{-a - n}^{b - n} \frac{\sin^2(\pi u)}udu - \int_{-a - m}^{b - m}\frac{\sin^2(\pi u)}udu\right) $$

Omitting the insides of the integrals to save space, we can break up the integrals like this: $$ \int_{-a - n}^{b - n} - \int_{-a - m}^{b - m} = \left[\int_{-a - n}^{-a - m} + \int_{-a - m}^{b - n}\right] - \left[\int_{-a - m}^{b - n} + \int_{b - n}^{b - m}\right] = \int_{-a - n}^{-a - m} - \int_{b - n}^{b - m}. $$ This leaves us with $$ \lim_{a,b \to \infty} \frac1{n-m} \left(\int_{-a - n}^{-a - m} \frac{\sin^2(\pi u)}udu - \int_{b - n}^{b - m}\frac{\sin^2(\pi u)}udu\right), $$ and both integrals clearly converge to $0$ as $a$ and $b$ increase.

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To justify the claim that the integrals "clearly" converge to $0$, I will prove it for the second integral. Without loss of generality, assume that $n > m$. Then, if $b > n$,

\begin{align} 0 < \int_{b - n}^{b - m}\frac{\sin^2(\pi u)}udu &\leq \int_{b - n}^{b - m}\frac1udu \\ &\leq \int_{b - n}^{b - m}\frac1{b - n}du \\ &= \frac{n - m}{b - n}. \end{align} As $b \to \infty$, $\frac{n-m}{b-n} \to 0$, which proves that the second integral converges to $0$. A symmetrical argument applies to the first integral.

Answer 4

Suffices to consider $$I=\int^\infty_{-\infty}\frac{\sin^2(\pi x)}{x(x-k)}\,dx=(-1)^k\int^\infty_{-\infty}\frac{\sin\pi x}{x}\frac{\sin\pi(x-k)}{x-k}\,dx$$ for $k\in\mathbb{N}$.

Notice that $$\mathcal{F}(\mathbf{1}_{[-1/2,1/2]})(t)=\int_{\mathbb{R}}e^{-i2\pi xt}\mathbf{1}_{[-1/2,1/2]}(x)\,dx=\frac{\sin \pi t}{ \pi t}$$ and $$ \mathcal{F}(\mathbf{1}_{[-1/2,1/2]})(t-k)=\int_{\mathbb{R}}e^{-i2\pi x(t-k)}\mathbf{1}_{[-1/2,1/2]}(x)\,dx=\frac{\sin \pi(t-k)}{\pi(t-k)}=\mathcal{F}(e^{2\pi ixk}\mathbf{1}_{[-1/2,1/2]})(t) $$

Since the Fourier transform is self-adjoint

\begin{align} \frac{(-1)^k}{\pi^2}I&=\int_{\mathbb{R}}\frac{\sin\pi t}{\pi t}\mathcal{F}(e^{2\pi ixk}\mathbf{1}_{[-1/2,1/2]})(t)\,dt=\int_{\mathbb{R}}\mathcal{F}\Big(\frac{\sin\pi t}{\pi t}\Big)(x) e^{2i\pi xk}\mathbf{1}_{[-1/2,1/2]}(x)\,dx\\ &=\int^{1/2}_{-1/2}e^{2\pi ixk}\,dx=\frac{e^{\pi ik}-e^{-\pi i k}}{2\pi i k}=\frac{\sin \pi k}{\pi k}=0 \end{align} since $k\in\mathbb{N}$.

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A more direct proof can be obtained without the use of Fourier analysis. Notice that $g:x\mapsto\frac{\sin^2\pi x}{x}$ is odd and locally integrable; furthermore, $x\mapsto g(x)g(x-k)$ is integrable. Then, for $a>0$ large enough \begin{align} I_a&=\int^a_{-a}\frac{\sin^2\pi x}{x(x-k)}\,dx=\frac{1}{k}\int^a_{-a}\Big(\frac{\sin^2\pi x}{x}-\frac{\sin^2\pi x}{x-k}\Big)\,dx\\ &=-\frac1k\int^{a-k}_{-a-k}\frac{\sin^2\pi x}{x}\,dx=-\frac1k\Big(\int^{-a+k}_{-a-k}+\int^{a-k}_{-a+k}\frac{\sin^2\pi x}{x}\,dx\Big)\\ &=-\frac1k\int^{-a+k}_{-a-k}\frac{\sin^2\pi x}{x}\,dx \end{align} finally, observe that $$\Big|\int^{-a+k}_{-a-k}\frac{\sin^2\pi x}{x}\,dx\Big|\leq \int^{a+k}_{a-k}\frac{dx}{x}=\log\big(\frac{a+k}{a-k}\big)\xrightarrow{a\rightarrow\infty}0$$ The conclusion follows by dominated convergence.