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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: $$ \int_{-\infty}^{\infty}\dfrac{\sin\left(x\right)}{x^{n}}\,\mathrm{d}x =A $$ .

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  • $\begingroup$ It's quite unclear how you're defining $m$ and $\mathbb{N}^*$. Care to clarify? Also, Wolfram Alpha seems to think that the integral is divergent for $n > 1$. $\endgroup$ – AlkaKadri Mar 13 '18 at 18:45
  • $\begingroup$ yes , it is divergent for n >1 $\endgroup$ – zeraoulia rafik Mar 13 '18 at 18:46
  • $\begingroup$ Probably the function is $\displaystyle \frac{\sin^n (x)}{x^n}$? $\endgroup$ – MathNovice Mar 13 '18 at 18:48
  • $\begingroup$ The Cauchy principal value is usually only well-defined if a function has simple poles on the real axis. This function has a pole of order $n-1$ on the real axis, which won't be a simple pole unless $n = 2$. $\endgroup$ – Michael Seifert Mar 13 '18 at 19:00
  • $\begingroup$ Actually, that's not accurate; the Cauchy principal value of $\int x^{-3} \, dx$ exists and is zero. However, the usual contour integration method fails in this case for $n \geq 3$, and I'm not sure why. I'll have to think about this some more. $\endgroup$ – Michael Seifert Mar 13 '18 at 19:29
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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.


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.)

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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:

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

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