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, 2019 at 15:13
  • 1
    $\begingroup$ Maybe use $$\int_0^\infty \frac{f(x)}x\mathrm dx=\int_0^\infty \mathcal{L}\{f\}(s)\mathrm ds$$? $\endgroup$
    – clathratus
    Feb 10, 2019 at 21:13

2 Answers 2


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, 2019 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, 2019 at 20:21

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.


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