I dont recall where, but I found this interesting identity a few years ago. It was shown in one of Victor Moll's papers about elliptic integrals.

Corollary 3.1. Let $f$ be an even function with period $a$. Then, $$\int_0^\infty \frac{f(x)}{x} \sin \bigl(\frac{\pi x}{a}\bigr) \,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \,\mathrm{d}x.$$

To prove this result the following lemma is applied

Lemma 3.1. Let $f$ be an odd periodic function of period $a$. Then $$ \int_0^{\infty} \frac{f(x)}{x} \,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} f(x) \cot\bigl( \frac{\pi x}{a} \bigr) \,\mathrm{d}x $$

I have no qualms about this lemma. Further, Victor writes that Corollary 3.1 follows from using the lemma on $f(x) = g(x) \sin (\frac{\pi x}{a})$ with the half-angle formula

$$ \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} $$

However, this is where my problems starts. Inserting in our values and using that $\cot x = 1/\tan x$

$$ \cot \bigl( \frac{\pi x}{a} \bigr) = \frac{1 + \cos(2\pi x/a)}{\sin(2\pi x/a)}$$

Inserting this into the lemma with $f = g \cdot \sin(\pi x/a)$ gives

$$ \int_0^{\infty} \frac{g(x)}{x} \sin \bigl( \frac{\pi x}{a} \bigr)\,\mathrm{d}x = \frac{\pi}{2} \int_0^{a/2} g(x) \sin \bigl( \frac{\pi x}{a} \bigr) \bigl[ 1 + \cos \bigl( \frac{2 \pi x}{a} \bigr) \bigr] \csc \bigl( \frac{2\pi x}{a} \bigr) \,\mathrm{d}x $$

I would really like the $\csc(x)$ and $\sin(x)$ to cancel out, however the extra factor of $2$ throws the entire calculation of. Any help proving the desired equality would be much obliged.

EDIT: Maybe the lemma is missing a 1/2 factor in the cotan argument?

  • 1
    $\begingroup$ This looks like a generealization of Lobachevsky's integral formula: math.stackexchange.com/questions/776903/… $\endgroup$ – Zacky Feb 10 '19 at 15:13
  • $\begingroup$ Maybe use $$\int_0^\infty \frac{f(x)}x\mathrm dx=\int_0^\infty \mathcal{L}\{f\}(s)\mathrm ds$$? $\endgroup$ – clathratus Feb 10 '19 at 21:13

robjohn has already given a complete derivation of both results. I would like only to point out the details which have prevented you from deriving the Corollary from the Lemma.

  1. There is either a misprint or a mistake in both Lemma and Corollary. The prefactor on the rhs of the equalities shoud be $\frac\pi a$ rather than $\frac\pi2$.

  2. As you quite correctly guessed there is a missing 1/2 factor in the cotan argument. However it is missing not in the lemma but in the derivation!

Indeed the function $$ f(x)=g(x)\sin\frac{\pi x}a, $$
where $g(x)$ is an even $a$-periodic function, is an odd $\color{red}{2a}$-periodic function. Therefore the Lemma 3.1 for this function reads: $$ \int_0^\infty g(x)\sin\frac{\pi x}a dx =\frac\pi{\color{red}{2a}}\int_0^\color{red}a g(x)\sin\frac{\pi x}a \cot\frac{\pi x}{\color{red}{2a}} dx. $$

PS. I have just looked into the cited paper and found that the authors used the correct prefactor.


Corollary 3.1

An even function with period $a$ means $f(a-x)=f(x)$. $$ \begin{align} \int_0^\infty\frac{f(x)}x\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x &=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+2ka}-\frac1{x+(2k+1)a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag1\\ &=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+2ka}+\frac1{x-(2k+2)a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag2\\ &=\sum_{k\in\mathbb{Z}}\int_0^af(x)\,\frac1{x+2ka}\,\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag3\\ &=\frac\pi{2a}\int_0^af(x)\cot\left(\frac{\pi x}{2a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag4\\[3pt] &=\frac\pi{2a}\int_0^af(x)\tan\left(\frac{\pi x}{2a}\right)\sin\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag5\\[3pt] &=\frac\pi{2a}\int_0^af(x)\,\mathrm{d}x\tag6\\ &=\frac\pi{a}\int_0^{a/2}f(x)\,\mathrm{d}x\tag7\\ \end{align} $$ Explanation:
$(1)$: $f(x+a)=f(x)$ and $\sin\left(\frac{\pi(x+a)}a\right)=-\sin\left(\frac{\pi x}a\right)$
$(2)$: $f(a-x)=f(x)$ and $\sin\left(\frac{\pi(a-x)}a\right)=\sin\left(\frac{\pi x}a\right)$ on the second term of the sum
$(3)$: write as a sum over $\mathbb{Z}$
$(4)$: use $(7)$ from this answer
$(5)$: $f(a-x)=f(x)$ and $\sin\left(\frac{\pi(a-x)}a\right)=\sin\left(\frac{\pi x}a\right)$
$(6)$: average $(4)$ and $(5)$
$(7)$: $f(a-x)=f(x)$

Lemma 3.1

An odd function with period $a$ means $f(a-x)=-f(x)$. $$ \begin{align} \int_0^\infty\frac{f(x)}x\,\mathrm{d}x &=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x+ka}\right)\,\mathrm{d}x\tag8\\ &=\sum_{k=0}^\infty\int_0^af(x)\left(\frac1{x-(k+1)a}\right)\,\mathrm{d}x\tag9\\ &=\frac12\sum_{k\in\mathbb{Z}}\int_0^af(x)\left(\frac1{x+ka}\right)\,\mathrm{d}x\tag{10}\\ &=\frac\pi{2a}\int_0^af(x)\cot\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag{11}\\ &=\frac\pi{a}\int_0^{a/2}f(x)\cot\left(\frac{\pi x}a\right)\,\mathrm{d}x\tag{12} \end{align} $$ Explanation:
$\phantom{1}(8)$: $f(x+a)=f(x)$
$\phantom{1}(9)$: $f(a-x)=-f(x)$
$(10)$: average $(8)$ and $(9)$ and write as a sum over $\mathbb{Z}$
$(11)$: use $(7)$ from this answer
$(12)$: $f(a-x)\cot\left(\frac{\pi(a-x)}a\right)=f(x)\cot\left(\frac{\pi x}a\right)$

  • $\begingroup$ (+1) Hi Rob. Happy New Year (Am I allowed to write that in February?). I don't see why this answer hasn't received several up votes. $\endgroup$ – Mark Viola Feb 10 '19 at 20:04
  • $\begingroup$ @MarkViola: Since January essentially evaporated, I see no reason why not. Happy New Year to you, as well. I never try to divine why answers get votes or not. $\endgroup$ – robjohn Feb 10 '19 at 20:21

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.