Evaluation of $\int_{-\infty}^{\infty} \frac{\sin(x)}{x^n} \,\mathrm{d}x $?

Question

The cauchy principal value of $ \displaystyle\int_{-\infty}^{\infty}\dfrac{\sin\left(x\right)}{x^n}\,\mathrm{d}x $ is defined as :

$$ \begin{cases} {\displaystyle A=0\,,\quad n\in \mathbb{N^*}} \\[2mm] {\displaystyle A=\frac{\pi\left(-1\right)^{m}}{\left(n - 1\right)!}\,,\qquad n \in 2\mathbb{N}+1\,,\quad n = 2m + 1} \end{cases} $$.

My question here is : How we can prove that: $$ \int_{-\infty}^{\infty}\dfrac{\sin\left(x\right)}{x^{n}}\,\mathrm{d}x =A $$ .

Answer 1

Well, we are looking at:

$$\mathcal{I}_{\space\text{n}}:=\int_{-\infty}^\infty\frac{\sin\left(x\right)}{x^\text{n}}\space\text{d}x\tag1$$

When $\text{n}$ is in the form of:

$$\text{n}=2\cdot\text{k}+1\tag2$$

Where $\text{k}\in\mathbb{N}$ (including $0$).

We know that $\frac{\sin\left(x\right)}{x^{2\cdot\text{k}+1}}$ is an even function, so we can write:

$$\mathcal{I}_{\space2\cdot\text{k}+1}=2\cdot\int_0^\infty\frac{\sin\left(x\right)}{x^{2\cdot\text{k}+1}}\space\text{d}x\tag3$$

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform we can write:

$$\mathcal{I}_{\space2\cdot\text{k}+1}=2\cdot\int_0^\infty\mathscr{L}_x\left[\sin\left(x\right)\right]_{\left(\text{s}\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x^{2\cdot\text{k}+1}}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}=$$ $$2\cdot\int_0^\infty\frac{1}{1+\text{s}^2}\cdot\frac{\text{s}^{2\cdot\text{k}}}{\Gamma\left(2\cdot\text{k}+1\right)}\space\text{d}\text{s}=\frac{2}{\Gamma\left(2\cdot\text{k}+1\right)}\cdot\int_0^\infty\frac{\text{s}^{2\cdot\text{k}}}{1+\text{s}^2}\space\text{d}\text{s}\tag4$$

Using this answer we can write:

$$\mathcal{I}_{\space2\cdot\text{k}+1}=\frac{2}{\Gamma\left(2\cdot\text{k}+1\right)}\cdot\frac{\pi\cdot\sec\left(\pi\cdot\text{k}\right)}{2}=\frac{\pi\cdot\sec\left(\pi\cdot\text{k}\right)}{\Gamma\left(2\cdot\text{k}+1\right)}\tag5$$

Answer 2

The answer below is incomplete; it uses a standard method for evaluating the principal value of a contour integral, but fails for reasons I can't quite understand. Any comments to fix it, or pointing out why it's fatally flawed, would be appreciated.

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The simplest way is to use the residue theorem. To apply this, consider the contour in the complex plane of the following form:

• I. From $z = +\epsilon$ to $z = R$, along the real axis
• II.From $z = R$ to $z = -R$, in a semi-circle centered at $z = 0$ in the upper half of the complex plane
• III. From $z = -R$ to $z = - \epsilon$, along the real axis
• IV. From $z = - \epsilon$ to $z = + \epsilon$, in a semi-circle centered at $z = 0$ in the upper half of the complex plane.

Consider the integral of the functions $f(z) = e^{iz}/z^n$ along this contour. Note that $\sin z/z^n$ is the imaginary part of $f(z)$.

In the limit $\epsilon \to 0$ and $R \to \infty$, the contributions from portions I and III of the contour become the Cauchy principal value of this integral. Moreover, since $f(z)$ converges to 0 as $z \to \infty$ in any direction in the upper half-plane, the integral over portion $II$ of the contour will go to zero in this limit. Finally, we note that this function does not contain any poles in the upper half-plane, so the integral around our contour will be zero. Putting this all together, we conclude that $$ \mathrm{P}\int_{-\infty}^\infty f(z) \, dz + \lim_{\epsilon \to 0} \int_{III} f(z) \, dz = 0. $$ So to find the Cauchy principal value, we must perform the integral around the small semi-circle in a clockwise direction. Along this curve, we have $z = \epsilon e^{i \phi}$, with $\phi$ running from $\pi$ to $0$. Thus, $$ \int_{III} f(z) \, dz = \int_{\pi}^0 f(\epsilon e^{i \phi}) \left( \epsilon i e^{i \phi} \right) \, d \phi. $$ Now, if we expand $f(z)$ in a Laurent series about $z = 0$, we have $$ f(z) = \sum_{m = 0}^\infty \frac{i^m z^{m-n}}{m!} $$ and so $$ f(\epsilon e^{i \phi}) \left( \epsilon i e^{i \phi} \right) = \sum_{m = 0}^\infty \frac{i^{m+1} \epsilon^{m-n+1}}{m!} e^{i \phi (m -n + 1)}. $$ Thus, $$ \int_{\pi}^0 f(\epsilon e^{i \phi}) \left( \epsilon i e^{i \phi} \right) \, d \phi = \sum_{m = 0}^\infty \left[ \frac{i^{m+1} \epsilon^{m-n+1}}{m!} \int_\pi^0 e^{i \phi (m - n + 1)} d \phi \right]. $$ For $m - n + 1 \neq 0$, this integral becomes $$ \int_\pi^0 e^{i \phi (m - n + 1)} d \phi = \begin{cases} 2 /i(m - n + 1) & m - n \text{ even} \\ 0 & m - n \text{ odd} \end{cases} \\ $$ while for $m - n + 1 = 0$, the integral is simply $-\pi$. Thus, we have $$ \int_{III} f(z) \, dz = -\frac{\pi i^{n}}{(n-1)!} + 2\sum_{m -n \text{ even}}^\infty \frac{i^{m} \epsilon^{m-n+1}}{m!(m-n+1)} $$ If $n$ is even, then all the terms in this expression will be real, and so the imaginary part will vanish. If $n$ is odd, then this expression will be purely imaginary, so we have $$ \Im \int_{III} f(z) \, dz = -\frac{\pi i^{n-1}}{(n-1)!} +2\sum_{m -n \text{ even}}^\infty \frac{i^{m-1} \epsilon^{m-n+1}}{m!(m-n+1)} $$ The limit of this integral, unfortunately, is not well-defined as $\epsilon \to 0$; all the terms in the sum with $m < n - 1$ will diverge, due to the factor of $\epsilon^{m-n+1}$. (This is ultimately due to the fact that the function does not have a simple pole at $z = 0$, but rather a pole of order $n-1$.)

However, if one uses a nonstandard principal-value prescription that allows us to simply discard all the $\epsilon$-dependent terms, one does in fact obtain $$ \mathrm{P}\int_{-\infty}^\infty \frac{\sin x}{x^n} \, dx = - \Im \int_{III} f(z) \, dz = \frac{\pi (-1)^{(n-1)/2}}{(n-1)!} $$ as desired. I don't think that this is the usual way that it's done, but it's possible that the reference you are working from defines things in a non-standard way. (It's also possible that I've made an error in my derivation, and I welcome corrections in the comments.)