Find integral of $\operatorname{sinc}(\pi x)$ over range $x=[0,1]$

Question

I want to find the area under a normalised $\operatorname{sinc}$ curve raised to the power of $2 k$ over the range $x=[0,1]$ where $k$ is a non-negative integer. In other words:

$$\int_0^1 \operatorname{sinc}^{2k}(\pi x) dx$$

It is well-established that $\int_0^\infty \operatorname{sinc}(\pi x) dx=\frac{1}{2}$. But I can't figure out how to take a definite integral within this more limited range, and with $\operatorname{sinc}$ raised to power $2k$ that I can then plug any chosen value of $k$ into. (I can use Mathematica to find integrals for a few specific values of $k$, but I can't find a general solution.)

Note that I don't really understand Fourier analysis, sorry. And things are likely to get hairy if I try to plug in a complex exponent because the definition that distinguishes $\operatorname{sinc}(\pi x)$ from $\frac{\sin (\pi x)}{\pi x}$ would be lost.

Is it possible to create this general definite integral? How?

EDIT:

Also, what happens in the limit as $k \rightarrow\infty$? Presumably the value of the integral tends to $0$... Perhaps that's why there's no general expression for it?

Answer 1

Frankly I do not really expect any closed form expressions for this integral since in the literature one actually defines the anti-derivative of the sinc function as a new special function, the sine integral https://en.wikipedia.org/wiki/Trigonometric_integral. However having said that it is always very easy to derive an infinite series that converges fast to those integrals in question. Firstly note that the $n$th primitive function of $1/x$ has a known form. The equation \begin{equation} \frac{d^n}{d x^n} y(x) = \frac{1}{x} \end{equation} is solved by \begin{equation} y(x) = \frac{x^{n-1}}{(n-1)!} \left(\log(x)-H_{n-1}\right) \end{equation} On the other hand the $n$th derivative of the sine function is also known; it alternates between a sine and a cosine depending on the remainder of division of $n$ by four. If we take this into account we can, by integrating by parts, easily write down the anti-derivative of the sinc function as an infinite series. Then all we need to do is to evaluate those series at unity and at zero and subtract the results. This leads to the following results: \begin{eqnarray} \int\limits_0^n \frac{\sin(\pi x)}{x} dx &= & \sum\limits_{m=0}^\infty \frac{(-n \pi)^{4 m+1}}{(4 m+1)!} H_{4m+1} - \sum\limits_{m=0}^\infty \frac{(-n \pi)^{4 m+3}}{(4 m+3)!} H_{4m+3}\\ \end{eqnarray} for $n=1,2,3,\cdots$. In addition to that using integration by parts we get: \begin{eqnarray} \int\limits_0^1 \frac{\sin(\pi x)^2}{x^2} dx &=& \pi \int\limits_0^2 \frac{\sin(\pi x)}{x} dx \\ \int\limits_0^1 \frac{\sin(\pi x)^3}{x^3} dx &=& \frac{3}{8} \pi^2 \left(-\int\limits_0^1 \frac{\sin(\pi x)}{x} dx+3\int\limits_0^3 \frac{\sin(\pi x)}{x} dx \right) \\ \int\limits_0^1 \frac{\sin(\pi x)^4}{x^4} dx &=& \frac{2}{3} \pi^3 \left(-\int\limits_0^2 \frac{\sin(\pi x)}{x} dx+2\int\limits_0^4 \frac{\sin(\pi x)}{x} dx \right) \\ \int\limits_0^1 \frac{\sin(\pi x)^5}{x^5} dx &=& \frac{5}{384} \pi^4 \left( 2 \int\limits_0^1 \frac{\sin(\pi x)}{x} dx - 81 \int\limits_0^3 \frac{\sin(\pi x)}{x} dx + 125 \int\limits_0^5 \frac{\sin(\pi x)}{x} dx \right)\\ \vdots \end{eqnarray} In general we have: \begin{eqnarray} \int\limits_0^1 \frac{\sin(\pi x)^{2 n}}{(\pi x)^{2 n}} dx &=& \frac{1}{2 (2n-1)!} \sum\limits_{p=0}^{2 n} \binom{2 n}{p} (-1)^p (n-p)^{2n-1} \int\limits_0^{2(n-p)} \frac{\sin(\pi x)}{\pi x} dx\\ \int\limits_0^1 \frac{\sin(\pi x)^{2 n-1}}{(\pi x)^{2 n-1}} dx &=& \frac{1}{2 (2n-2)!} \sum\limits_{p=0}^{2n-1} \binom{2n-1}{p} (-1)^p (n-p-1/2)^{2n-2} \int\limits_0^{2n-2 p-1} \frac{\sin(\pi x)}{\pi x} dx \end{eqnarray}