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Thanks for the A2A. I use the following definition of the Fourier transform of [math]f(x) = 1/\cosh{ax}[/math]:[math]\begin{align}\mathcal{F}[f](k) = \int_{-\infty}^\infty\,\frac{e^{ikx}}{\cosh{ax}}\,dx = 2\int_{-\infty}^\infty\,\frac{e^{ikx}}{e^{ax} + e^{-ax}}\,dx\,,\end{align}[/math]omitting WolframAlpha's normalization factor of [math]1/\sqrt{2\pi}[/math]. My answer follows a solution procedure outlined at Fourier transform of 1/cosh by Felix Marin, filling in a number of steps that are missing there.First, it is clear from the evenness of [math]\cosh{ax}[/math] that [math]a[/math] can be replaced by [math]|a|[/math] without loss of generality, that is, [math]\cosh{ax} = \cosh{(|a|x)} = (e^{|a|x} + e^{-|a|x})/2[/math]. Making a change of variables to [math]t[/math][math] = e^{|a|x}[/math], it then follows that as [math]x \to -\infty[/math], [math]\,[/math][math]t[/math][math] \to 0[/math], and as [math]x \to \infty[/math], [math]\,[/math][math]t[/math][math] \to \infty[/math]. Note also that [math]e^x = t^{1/|a|}[/math]. Since [math]dt = |a| e^{|a|x} dx,[/math] we have [math]dx = dt/|a|t[/math], and the transform reduces to[math]\begin{align}\mathcal{F}[f](k) &= 2\int_{-\infty}^\infty\,\frac{e^{ikx}}{e^{|a|x} + e^{-|a|x}}\,dx\\ &= 2\int_{0}^\infty\,\left(\frac{t^{-ik/|a|}}{t + 1/t}\right)\,\frac{dt}{|a|t}\\ &= \frac{2}{|a|}\int_{0}^\infty\,\frac{t^{-ik/|a|}}{t^2 + 1}\,\,dt\,.\end{align}[/math]Make another change of variables to [math]u = t^2[/math], hence [math]t[/math][math] = u^{1/2}[/math], leaving the limits unchanged, and note that [math]du = 2t dt = 2 u^{1/2} dt[/math], so that [math]dt = u^{-1/2}du/2[/math], bringing the transform to[math]\begin{align}\mathcal{F}[f](k) = \frac{1}{|a|}\int_{0}^\infty\,\frac{u^{-ik/2|a|-1/2}}{u + 1}\,\,du\,. \tag{1}\end{align}[/math]To evaluate the remaining integral, we apply the residue theorem to the integral[math]\begin{align}I(k) = \oint_{\mathcal{C}}\,\frac{z^{-ik/2|a|-1/2}}{z + 1}\,\,dz\,,\end{align}[/math]where [math]\mathcal{C}[/math] is the contour illustrated in the following Figure:In what follows, I relied on a problem solved in the Schaum's Outline on Complex Variables (2nd edition, Chapter 7, page 223, Problem 7.20). The integrand of [math]I(k)[/math] has branch points at [math]z = 0[/math] and [math]\infty[/math], so we choose the positive real axis as the branch line where segments [math]AB[/math] and [math]GH[/math], shown separated for convenience, are actually coincident with the positive [math]x[/math]-axis. It has a simple pole at [math]z = -1 = e^{i\pi}[/math] inside [math]\mathcal{C}[/math] with residue[math]\begin{align}\lim_{z\to -1}(z+1)\frac{z^{-ik/2|a|-1/2}}{z+1} = e^{i\pi(-ik/2|a|-1/2)} = e^{\pi k/2|a| - i\pi/2} = -ie^{k\pi/2|a|}\,,\end{align}[/math]since [math]e^{-i\pi/2} = -i[/math]. Writing the integral as the sum of integrals over the four paths [math]AB[/math], [math]BDEFG[/math], [math]GH[/math], and [math]HJA[/math], which by the residue theorem must be equal to [math]2\pi i[/math] times the residue, we have[math]\begin{align}\int_{AB} + \int_{BDEFG} + \int_{GH} + \int_{HJA} = 2\pi e^{k\pi/2|a|}\,.\end{align}[/math]On the large circle [math]BDEFG[/math], we have [math]z = Re^{i\theta}[/math], while on the small circle [math]HJA[/math], we have [math]z = \epsilon e^{i\theta},\,\, 0<\epsilon<1[/math]. The tricky (at least to my mind) point is to notice that on [math]GH[/math], although corresponding points on this line segment and the segment [math]AB[/math] are the same distance [math]x[/math] from the origin, they do not have the same value of [math]z[/math], since the argument of [math]z[/math] has increased by [math]2\pi[/math] in going around the large circle. So we must set [math]z = xe^{2\pi i}[/math] on [math]GH[/math] rather than simply [math]x[/math]. The integrals may then be written as[math]\begin{multline}\int_{\epsilon}^{R}\,\frac{x^{-ik/2|a|-1/2}}{x + 1}\,dx + \int_{0}^{2\pi}\,\frac{(Re^{i\theta})^{-ik/2|a|-1/2}}{Re^{i\theta} + 1}\,R i e^{i\theta}\,d\theta \\+ \int_{R}^{\epsilon}\,\frac{(x e^{2\pi i})^{-ik/2|a|-1/2}}{xe^{2\pi i} + 1}\,e^{2\pi i}\,dx + \int_{0}^{2\pi}\,\frac{(\epsilon e^{i\theta})^{-ik/2|a|-1/2}}{\epsilon e^{i\theta} + 1}\,\epsilon i e^{i\theta}\,d\theta = 2 \pi e^{k\pi/2|a|}\,.\end{multline}[/math]In the limits [math]R \to \infty[/math] and [math]\epsilon \to 0[/math], the second and fourth integrals are easily seen to vanish, while the third integral can be simplified, leaving[math]\begin{align}\int_{0}^{\infty}\,\frac{x^{-ik/2|a|-1/2}}{x + 1}\,dx + \,e^{k \pi/|a| - i\pi}\int_{\infty}^{0}\,\frac{x^{-ik/2|a|-1/2}}{xe^{2\pi i} + 1}\,e^{2\pi i}\,dx = 2 \pi e^{k\pi/2a}\,,\end{align}[/math]or, using [math]e^{-i\pi} = -1[/math], and reversing the order of integration on the second integral (thus changing its sign), we have[math]\begin{align}\left(1 + [/math][math]e[/math][math]^{ [/math][math]k[/math][math] \pi/|a|}\right)\int_{0}^{\infty}\,\frac{x^{-ik/2|a|-1/2}}{x + 1}\,dx = 2 \pi e^{k\pi/2|a|}\,.\end{align}[/math]Replacing the dummy integration variable [math]x[/math] by [math]u[/math] , and comparing to equation (1), we see that this is simply[math]\begin{align}\left(1 + e^{k \pi/|a|}\right)\,|a|\,\,\mathcal{F}[f](k) = 2 \pi e^{k\pi/2|a|}\,,\end{align}[/math]from which we obtain the Fourier transform:[math]\begin{align}\mathcal{F}[f](k) = \frac{2 \pi e^{k\pi/2|a|}}{|a|(1 + e^{k \pi/|a|})} = \frac{\pi}{\dfrac{|a|(e^{k\pi/2|a|} + e^{-k\pi/2|a|})}{2}} = \frac{\pi}{|a|\,\cosh{(k\pi/2a)}}\,.\end{align}[/math]Multiplying by WolframAlpha's normalization factor [math]1/\sqrt{2\pi}[/math] brings this into agreement with the result posted in David Goldsmith's answer.

I happened to do some research on this matter for another answer, so I have the resources at hand to answer this one.The problem, however, is that there are SO many languages, and so much data, that it would be a really crazy task to put them all together.Sounds like a job for me then!I selected 27 of the most widely known and spoken languages that use common types of alphabets. The first row for each language represents order of frequency of all characters, accented ones separated, and the second row has the order of letter frequency when you group accented letters into their original letter, and when you split digraphs into their original letters (note: I made an exception with æ and œ and did not split them).Afrikaans:e a i n r s o t d l k g v m p u b w f h j c ê ë z öe a i n r s o t d l k g v m p u b w f h j c zAlbanian:ë e t z i a r n m u s x o k j sh y p l d b f h q g rr v dh gj ll nj th c çe t z i r a n m s u j x l o k h d y p g b f q v cCatalan:e a s r l t i n o u m d c p v b q g f h é x ó j è à ò í y ç k ú z ü w ïe a s r i l t n o u m d c p v b q g f h x j y k z wCzech:a e o n i t v s r l d k m u p í z j h ě y ý á b c ž š é č ř ůe i a o n s t v r l d k m u z y p j h c bDanish:e r n t a i d s l o g k m f v b u p h å ø æ j y c w z x qe r n a t i d s o l g k m f v b u p h æ j y c w z x qDutch:e n a t i r o d s l g v h k m u b p w j z c f x y qe n a t i r o d s l g v h k m u b p w j z c f x y qEnglish:e t a o i n s h r d l c u m w f g y p b v k j x q ze t a o i n s h r d l c u m w f g y p b v k j x q zEsperanto:a i e o n l s r t k j u d m p v g f b c ĝ ĉ ŭ z ŝ h ĵ ĥa i e o n s l r t k u j d m p v g c f b z hEstonian:i a e s t u o k l n d r g m v j h õ p b ü ä ö fi a e s t u o k l n d r g m v j h p b fFinnish:a i n t e s l o u k ä m r v j h p y d ö g b c f w z x qa i n t e s l o u k m r v j h p y d g b c f w z x qFrench:e s a i t n r u o l d c m p v é q f b g h j à x z è ê y ç w ù â k î ô œ ë ïe a s i t n r u o l d c m p v q f b g h j x z y w k œGerman:e n s r i a t d h u l g c o m w b f k z ü v p ä ö ß j y x qe n s a r i t u d h l o g c m w b f k z v p ß j y x qGreek:α o ε ι τ σ ν η ρ π υ k μ λ ω γ δ χ θ φ β ξ ζ ψα o ε ι τ σ ν η ρ π υ k μ λ ω γ δ χ θ φ β ξ ζ ψHawaiian:a k o i n e u h l m p wa k o i n e u h l m p wHungarian:e a t n l s k é i m o á g r z v b d sz j h gy ő ö ny ly ü ó f p í u cs ű c ú zse a t n l s o i k m z g r v b d j y h u f p cIcelandic:a r n i e s t u l ð g m k f v o h á d í þ j b ó y æ p ö é ú ý xa i r n e d s u t l g m o k f v h þ j y b æ p xIndonesian:a n e i t k d r u m s g l h b p y o j c w f v z x é qa n e i t k d r u m s g l h b p y o j c w f v z x qItalian:e a i o n l r t s c d p u m v g z f b h à q è ù w ì y j ka e i o n l r t s c d u p m v g z f b h q w y j kLatin:i e a u t s r n o m c l p d b q g v f h x y zi e a u t s r n o m c l p d b q g v f h x y zNorwegian:e n t r a i s d l o g k m v h u p f å b j ø y c w æ z x qe n t r a i s d l o g k m v h u p f b j y c w æ z x qPolish:a i e o n w r s z c d y k l m t p ł u j b g ó ę h ś ć ż ą ń f ź v xa i e o n s w z r l c d y k m t p u j b g h f v xPortuguese:a e o s r i d m n t c u l p v g q b f h ã ô â ç z ê j é ó x ú í á à w ü k ya e o s r i d m n c t u l p v g q b f h z j x w k yRomanian:a i e l u r t n c o s m p d v f î ș b ă g z ț j h xa i e l u r t n c o s m p d v f b g z j h xRussian:о е а и н т с л в р к м д п ы у б я ь г з ч й ж х ш ю ц э щ ф ё ъо е и а н т с л в р к м д п ы у б я ь г з ч й ж х ш ю ц э щ ф ъSpanish:e a o s r n i d l t c m u p b g v y q ó í h f á j z é ñ x ú w ü ke a o s n i r d l t c m u p b g v y q h f j z x w kSwedish:e a n r t s i l d o m k g v h f u p ä b c å ö y j x w z qa e n r t s i o l d m k g v h f u p b c y j x w z qTurkish:a e i n r l k d ı m y t u s o b ü ş z c g h ç ğ v p ö f ji a e n r l k d u s o m y t b c g z h v p f jHere's how the whole thing looks put together. The numbers represent the ranking of the letter. Click on the image if you want to see the non-blurry version.A few interesting notes and observations:Nordic languages seem to have more consonant weight, compared to most of the other languages.Hungarian doesn't have the letter y by itself, only in the "gy", "ny", and "ly" digraphs.The lowest ranking for the letter a is in Albanian (6th).The lowest ranking for the letter e is 5th (Icelandic, Finnish)The ranking of the letter c ranges from 9th (Romanian) to 25th (Albanian)Estonian is almost the only language that doesn't use the letter f, except in loan words. Strange, huh?English has an unusually high ranking for the letter h (8th) - the average for the other languages is 18th and the second highest is 13th. No wonder though, considering words such as the, hello, hi, he/him/his, she/her/hers, when, how, what, where, who, which, why.The usage of the letter k varies from sixth place (Indonesian) to last (Spanish, Italian)On average, o is somewhere around 8th, but it varies from 3rd (Spanish, Portuguese) to 18th (Indonesian)Latin and French have the highest use of the letter q (16th).The letter t is also very variable, from second (Albanian, English), to 16th (Polish).The same goes with u, which ranks as high as 4th in Latin, and as low as 20th in Hungarian.Czech uses the letter v a lot more than all other languages. It's the 8th most used, as compared to the overall average of 20.A similar thing happens with w in Polish, where it's 7th, while the overall average is 22ndAlbanian wins at the usage of the letter z, it ranks 3rd, very far from the overall average of 21st (although Polish is not too far away with z ranked 8th).And the last interesting bit of info: if you add together the weight of each letter for all languages, you get the overall weight of every letter. That would arrange the (most common) alphabet into:e a i n r t s o l d u m k g c p v h b z f j y w q xSo in the end, as it was clearly obvious and visilble, E takes the lead, followed closely by A, while U is the least popular vowel. Sorry U.At the same time, N is the most frequently used consonant. Huh. I did not see that coming.Sources from where I dug up the letter frequencies (which you can blame for any errors, I mean, I'm not completely sold on z being the third most used letter in Albanian, but hey):WikipediaCryptogramSttmediaLater edit: For fun, I decided to test the letter frequencies of a lengthy answer. So I went into my feed and picked the longest one I could find with a quick search. Did the letter counts, and the results are below!e (378) / t (323) / a (291) / i (251) / o (247) / n (233) / s (223)r (171) / h (149) / l (130) / c (112) / u (110) / d (108) / w (81)m (77) / y (72) / f (63) / g (59) / b (44) / p (41) / k (37)v (36) / j (9) / x (7) / q (3) / z (2)And here it is (first line), in comparison with the general reorganized English alphabet (second line):e t a i o n s r h l c u d w m y f g b p k v j x q ze t a o i n s h r d l c u m w f g y p b v k j x q zQuite similar!Second edit: I know there are a few mistakes regarding some letters, but unfortunately I deleted the file I used to calculate the frequencies, and it would be a huge pain to re-create it. Maybe some day.

This is a special case of a classic integral whose answer is comprised of the zeta and gamma functions. More precisely, we consider the more general integral (with [math]n > -1[/math])[math]\displaystyle \int_0^{\infty} \frac{x^n}{e^x - 1} \, dx. \tag*{}[/math]First, we multiply the numerator and denominator by [math]e^{-x}:[/math][math]\displaystyle \int_0^{\infty} \frac{x^n e^{-x}}{1 - e^{-x}} \, dx. \tag*{}[/math]Why did we do this? Now, we can legally use the infinite geometric series (with ratio [math]e^{-x}[/math], which is less than 1 in magnitude for all [math]x > 0[/math]. This yields[math]\displaystyle \int_0^{\infty} x^n e^{-x} \cdot \sum_{k = 0}^{\infty} (e^{-x})^k \, dx. \tag*{}[/math]Interchanging the order of summation and the integral sign, we can rewrite this as[math]\displaystyle \sum_{k = 0}^{\infty} \int_0^{\infty} x^n e^{-(k+1)x} \, dx. \tag*{}[/math]In order to make the integral look more like the gamma function, we make the substitution [math]t[/math][math] = (k+1)x[/math]. This gives us[math]\displaystyle \sum_{k = 0}^{\infty} \int_0^{\infty} \Big(\frac{t}{k+1}\Big)^n e^{-t} \cdot \frac{dt}{k+1} = \sum_{k = 0}^{\infty} \frac{1}{(k+1)^{n+1}} \int_0^{\infty} t^n e^{-t} \, dt. \tag*{}[/math]By definition of the gamma function, the last integral equals [math]\Gamma(n+1)[/math]. Next, we re-index the summation to begin at 1; this yields[math]\displaystyle \Gamma(n+1) \cdot \sum_{k = 1}^{\infty} \frac{1}{k^{n+1}}. \tag*{}[/math]However, this latter sum is the Riemann zeta function [math]\zeta(n+1)[/math]. Hence, we have found that[math]\displaystyle \int_0^{\infty} \frac{x^n}{e^x - 1} \, dx = \Gamma(n+1) \zeta(n+1). \tag*{}[/math]Back to our original integral; letting [math]n = 1[/math] yields[math]\displaystyle \int_0^{\infty} \frac{x}{e^x - 1} \, dx = \Gamma(2) \zeta(2) = 1! \cdot \frac{\pi^2}{6} = \frac{\pi^2}{6}. \tag*{}[/math]