#Singular Log-Sine Integral from 0 to 1

72 messages · Page 1 of 1 (latest)

hybrid stumpBOT
frozen hearth
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so i made it first symmetric

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mine is coming 0

plucky pier
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that's nice

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so just numerical integration is enough

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that's also what wolfram alpha showed

frozen hearth
unique osprey
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,w euler-mascheroni constant

amber iceBOT
unique osprey
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ah

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looked suspiciously close lol

tender locust
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@vivid fox this is it babe

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@vivid fox

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Don’t mind me 😭

unique osprey
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what makes you think that there will be a closed form?

unique osprey
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i still gave it a shot with the first substitution you said and arrived to a series representation of this value

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$$\lim_{N\to\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} \underbrace{\int_1^N \frac{\ln(v) v^{2k}}{v-1} dv}_{I_k}$$

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but i'm 99% sure you can't do anything to simplify I_k

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not even for any specific k

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wait

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typo

amber iceBOT
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artemetra

474467399207223307.png
unique osprey
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major typo lol

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should be good now

unique osprey
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there is a way through the dilogarithm

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,w integral (ln(v))/(v-1)

amber iceBOT
unique osprey
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^ if you evaluate that on from 1 to N that's I_0

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,w integral (ln(v) v^2)/(v-1)

amber iceBOT
unique osprey
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here's I_1

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,w integral (ln(v) v^10)/(v-1)

amber iceBOT
unique osprey
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here's I_5

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i'm pretty sure what i happening here is wolframalpha spamming integration by parts

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first integrating ln(v)/(v-1) to get the dilogarithm form and then just dealing with a regular rational integral

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brb

unique osprey
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okay it's late here and i can't write the complete proof but you'll have to trust me on this lol

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$$I_{k} = \left.\left(-\operatorname{Li}2 (1-v) - \sum{j=1}^{2k} \left[\frac{v^j}{j^2} - \frac{1}{j} v^j \ln(v)\right]\right)\right|_{v\to 0}^{v=1-N}$$

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something like this

amber iceBOT
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artemetra

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unique osprey
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no guarantee that this makes any sense but i am pretty sure it does lol

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the really cool thing is that as v->0, all the terms in the sum vanish

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so we simply have $I_k = -\operatorname{Li}_2(N)$

amber iceBOT
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artemetra

474467399207223307.png
unique osprey
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wait no that makes no sense

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yeah i'm wrong, it's v->1 not v->0

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$$I_{k} = \left.\left(-\operatorname{Li}2 (1-v) - \sum{j=1}^{2k} \left[\frac{v^j}{j^2} - \frac{1}{j} v^j \ln(v)\right]\right)\right|_{v=1}^{v=1-N}$$

amber iceBOT
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artemetra

474467399207223307.png
unique osprey
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don't know where you can if you can take this much further

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at v=1 all the ln terms are 0 so that's nice ig

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but other than that this is probably the closest you'll get to a closed form

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if i write it as one expression..

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$$\lim_{N\to\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} \left.\left(-\operatorname{Li}2 (1-v) - \sum{j=1}^{2k} \left[\frac{v^j}{j^2} - \frac{1}{j} v^j \ln(v)\right]\right)\right|_{v=1}^{v=1-N}$$

amber iceBOT
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artemetra

474467399207223307.png
unique osprey
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take this with a grain of salt though

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it's 1 am here

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@dreamy galleon this is probably the farthest you can get it to be, although i likely made a mistake somewhere along the way. if this doesn't suffice, i recommend you post this on math stack exchange

plucky pier
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oh yea, how many divisions did you use for Simpson's rule @dreamy galleon

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i see

unique osprey
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i thought so too but it does not, ghe factorial outgrows the exponential. this is why the limit is outside the whole thing: you initially pick a huge N and then evaluate it

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but again this is probably VERY numerically unstable so uh yeah you have a point

glacial dagger
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interesting

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have you tried u=1/x

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duh obviously you've tried that

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I'll try it tmrw morning 👍

unique osprey
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it uses Meijer G function and the generalized hypergeometric function (that's a mouthful)