5
$\begingroup$

I have found the following complicated integral in Table of Integrals, Series and Products (page 469; No. 37); the interesting thing about this integral is that for arbitrary parameters $p_k,a_k>0$ and arbitrary natural number $n$ it has the value $\frac{\pi}{2}$:

$$\int_0^{\infty} \frac{\sin(bx)\sin(x)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \text{d} x= \frac{\pi}{2}$$

A large integral that has only one value when integrated over the interval $[0, \infty]$, but how I can prove this interesting fact? Series expansion in the trigonometric functions does not make sense, I think. Can I use induction for proving this identity?


The previous link to a PDF of the book no longer works.

$\endgroup$
  • $\begingroup$ You forgot one condition : $b> \sum_{k=1}^n a_kp_k$. $\endgroup$ – user37238 Mar 4 '15 at 9:58
  • $\begingroup$ Have you looked on the reference book indicated in the pdf you've linked? $\endgroup$ – user37238 Mar 4 '15 at 10:03
  • $\begingroup$ no, I haven't yet. $\endgroup$ – kryomaxim Mar 4 '15 at 10:06
  • 2
    $\begingroup$ Please make titles informative and objective. $\endgroup$ – Pedro Tamaroff Mar 4 '15 at 10:32
  • 1
    $\begingroup$ Hint: Because your integrand is even, you can use contour integration. The only contribution to the integral will be from $0$ and yields a constant.... $\endgroup$ – tired Mar 4 '15 at 14:40
1
$\begingroup$

EDIT: My answer was more complicated than it needed to be, so I made it simpler.


This is basically what is called a Borwein integral.

As was mentioned by user37238 in the comments, there is an additional condition stated in the table, namely $b> \sum_{k=1}^n a_{k}p_{k} $.

But this condition is not quite correct. It should be $$b \ge 1 + \sum_{k=1}^n a_{k}p_{k}.$$

Presumably we also want the $p_{k}$'s to be positive integers; otherwise, the integral is not well-defined.


Consider the complex function $$f(z) = \frac{e^{ibz}\sin(z)}{z^2} \prod_{k=1}^n \cos^{p_k}(a_{k}z).$$

What we want to show is that if $b \ge 1 + \sum_{k=1}^n a_{k}p_{k}$, then $$ \text{PV} \int_{-\infty}^{\infty} \frac{e^{ibx}\sin(x)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \, d x= i \pi \operatorname{Res}[f(z), 0] = i \pi (1) =i \pi. $$

(The result then follows if we equate the imaginary parts on both sides of the equation.)

To show this, all we need to do is argue that the magnitude of $$g(z) = e^{ibz} \sin(z) \prod_{k=1}^n \cos^{p_k}(a_kx)$$ is bounded in the upper half-plane if $b \ge 1 + \sum_{k=1}^n a_{k}p_{k}$.

It will then follow immediately from the estimation lemma that $\int f(z) \, dz$ vanishes along the upper half of the circle $|z|=R$ as $R \to \infty$.


If the upper half-plane, the magnitude of $e^{i \alpha z}, \, \alpha \ge 0$, is less than or equal to $1$.

The function $g(z)$ can be expressed as $$e^{ibz}\left(\frac{e^{iz}-e^{-iz}}{2i} \right) \prod_{k=1}^{n} \left(\frac{e^{i a_{n}z}+e^{-ia_{n}}z}{2} \right)^{p_{n}}.$$

If we expand, we get a linear combination of $2^{n+1} $ exponential functions of the form $e^{i \beta_{k} z}$, where $ \beta_{k} \in \mathbb{R}$.

The $\beta_{k}$'s range from $b+1+\sum_{k=1}^{n}a_{k}p_{k}$ to $b-1-\sum_{k=1}^{n}a_{k}p_{k} $.

If $b \ge 1+ \sum_{k=1}^{n} a_{n}p_{n}$, then the magnitude of $g(z)$ is bounded in the upper half-plane since all of the exponential functions are of the form $e^{i \alpha z}, \, \alpha \ge 0$.


A slight modification of the above argument shows that $$\int_{0}^{\infty} \frac{\sin (bx) \sin(rx)}{x^2} \prod_{k=1}^n \cos^{p_k}(a_kx) \, d x= \frac{\pi r}{2}, \quad b,r,a_{k} >0, \ p_{k} \in\mathbb{Z_{>0}}, \ b \ge r+ \sum_{k=1}^{n} a_{k}p_{k}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.