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The integral is: $$ \int_{ }^{ }\frac{\left(\arccos\left(x\right)\sqrt{1-x^{2}}\right)^{-1}}{\log_{e}\left(1+\frac{\arcsin\left(2x\sqrt{1-x^{2}}\right)}{\pi}\right)}dx. $$ I can't write all my work using latex, but the following links contain images with my work: (I'm aware I forgot the +c)

https://cdn.discordapp.com/attachments/882113354750128253/1157144865881215096/IMG_7749.jpg?ex=65183362&is=516e1e2&hm=b9ebb214b99704774ea0cefb88f60ced8cbc6d0bb995eb1204bd4f933951b09b&

https://cdn.discordapp.com/attachments/882113354750128253/1157144866233524304/IMG_7750.jpg?ex=65183362&is=616e1e2&hm=facfd97c578e407c101d9caac962a99295b650622cc337dfda91d905c6867058&

I'm aware my inductive proof is not perfect in terms of syntax and mathematical standards, but proving the relationship between $I_n$ and $I_{n+1}$ and proving that $I_1$ follows this relationship allows me to prove the conjecture upholds. Giving the following solution:

[edit: my original answer missed the (-) outside the integral, following is the same answer I got but negative]

$-\frac{1+\frac{2}{\pi}\cos^{-1}\left(x\right)}{\ln\left(1+\frac{2}{\pi}\cos^{-1}\left(x\right)\right)}\cdot\sum_{n=0}^{\infty}\frac{n!}{\left(\ln\left(1+\frac{2}{\pi}\cos^{-1}\left(x\right)\right)\right)^{n}}+\frac{\pi}{2}\ln\left(\ln\left(1+\frac{2}{\pi}\cos^{-1}\left(x\right)\right)\right) + c$

Not sure if I made any other small mistakes so let me know if this result is correct. My problem arises when Wolfram alpha and integral calculator both say it has no solution in elementary functions, and gives weird solutions for definitive integrals of this function. So I'm not sure If my answer is correct.

How would this integral work if it was definite? would the summation not always diverge to infinity? does this mean any definite integral would be infinite or would it be infinity-infinity $= 0$? This does not allign with the graph, and neither does computer calculated results.

Wolfram alpha and other integral calculators say it doesn't have an antiderivative in elementary functions. Does this not count because of the summation? Aren't they also elementary? Is it possible to find a function or at least a value for the definitive integral? for example, is it possible to find the general form of the definite integral from a to b such that $1 > b > a > 0 ,$(graph is not defined for $x=0$, $x \ge 1$, $x \le -1$) because the graph would imply this value converges.

Using integral calculator however(which also wasn't able to give a closed form answer or an answer to the indefinite integral), the approximated integral from 0.1 to 0.9 was around −4.73170536005842, which doesnt make much sense since the graph is positive between 0.1 and 0.9, as far as I'm aware. following is a screenshot of the graph between the limits:

https://cdn.discordapp.com/attachments/870017948478210118/1157398057789440072/Screenshot_2023-09-29_at_3.25.58_PM.png?ex=65187670&is=651724f0&hm=3aa57588a8f2adc6e244eaa0531bbf5b9cb50305f6f3e1a2649bce4cfdc4a3b6&

Appears to have a positive area, but the integral is evaluated as negative (the upper bound is lower than the lower bound)

Any help?

Edit: After some research, I see that the integral of (e^u)/u is Ei(u), the exponential integral. The Wikipedia page however did not have the same summation I had, it was off by a (-), the coefficient of the denominator of the summation fraction. I still do not fully understand how it works, or why the definite integral does not have a value for limits where the function is defined. I do however understand that Ei is not an elementary function, but I don't see why this causes contradictions with the definite integral.

Still not sure how a definite integral would work for such a solution, as the summation seems to diverge for all values of u (since factorials grow faster than exponentials), but this would contradict the graph of the function, which seems to have a definite area under it for certain points.

This also doesn't line up with Wolfram alphas solution, which is:

$\int_{ }^{ }\frac{e^{2u}}{u}=\ Ei\left(2u\right)$

Which assuming $\int_{ }^{ }\frac{e^{u}}{u}=\ Ei\left(u\right) = \frac{e^{u}}{u}\sum_{n=0}^{\infty}\left(\frac{n!}{u^{n}}\right) = Ei(u)$, isn't consistent with my answer

Even Ei(2x) has contradictions for the numerical approximation of the definitive integral from 0.1 to 0.9, giving a negative answer despite the graph being strictly positive for x(0.1,0.9)

Any help?

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  • $\begingroup$ No idea, but the expression being integrated is kinda gross, so it wouldn't surprise me one bit if it couldn't be written in terms of elementary functions. $\endgroup$ Commented Sep 29, 2023 at 20:25
  • $\begingroup$ @AdamRubinson Is my solution at least correct? $\endgroup$
    – Mihindu
    Commented Sep 29, 2023 at 21:01
  • $\begingroup$ This is very similar to a question posted yesterday math.stackexchange.com/questions/4777398/… and on mathoverflow on another account mathoverflow.net/questions/455549/… $\endgroup$
    – whpowell96
    Commented Sep 29, 2023 at 21:17
  • $\begingroup$ Additionally, the LaTeX (as well as the image in the aforementioned previous posts) have a $\sin$ instead of a $\sin^{1-}$ in the denominator. Which is correct? $\endgroup$
    – whpowell96
    Commented Sep 29, 2023 at 21:21
  • $\begingroup$ @whpowell96 Yes that's mine, the questions are different though. The sin at the bottom made the integral impossible, to my knowledge. changing it to arcsin made it possible, and in my attempts to solve it, this is what i got, and im trying to find a solution to the contradictions between my answer, the graph, and integral calculator. I then made this same post on math overflow but someone said the question type is better suited for math stack exchange and to post it there, so I made it here. $\endgroup$
    – Mihindu
    Commented Sep 29, 2023 at 21:35

1 Answer 1

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My attempt

     Assuming[ 0 < x < 1 , \[DifferentialD]x/(
ArcCos[x] Sqrt[1 - x^2] Log[1 + 1/\[Pi] ArcSin[2 x Sqrt[1 - x^2]]]) //.
{ArcCos[x] :>  u, 
 \[DifferentialD]x :>  \[DifferentialD]u/D[ArcCos[x], x], 
 x :> Cos[u], 
 Sqrt[1 - Cos[u]^2] -> Abs[Sin[u]],
 ArcSin[2 Abs[Sin[u]] Cos[u]] :>  2u Sign[Sin[u]]} // 
 PowerExpand //  FullSimplify] 

$$-\frac{du}{u \ \log \left(\frac{2 u \ \text{sgn}(\sin (u))}{\pi }+1\right)}$$

Mathematicas substitution yields a similar form

    Assuming[ 0 < u < \[Pi]/2 && Cos[2 u] > 0, 
     FullSimplify[
       IntegrateChangeVariables[
         Assuming[-1 < x < 1, 
          Inactive[Integrate][ 1/(  ArcCos[x] Sqrt[1 - x^2] *
               Log[1 + 1/\[Pi] ArcSin[2 x Sqrt[1 - x^2]]]), x]], 
          u,  u == ArcCos[x]]]  ] 

$$\text{Integrate}\left[\frac{1}{u \log \left(\frac{\pi }{2 u+\pi }\right)},u,\text{Assumptions}\to \text{True}\right]$$

with no known result for,

$$-\int \frac{1}{u \log (u+1)} \, du$$

as it seems, since the substitution of the Log yields

$$\int \frac{1}{t \left(1-e^t\right)} \, dt$$

with no result either.

Are your integral expressions results of substitutions of an easier one, perhaps?

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  • $\begingroup$ @Rolad F I realise I made a mistake. My first substitution was the same as yours, i think (x = cos(u)). This gave $\int_{ }^{ }\frac{-u}{\ln\left(1+\frac{2}{\pi}u\right)}$. Then I used the substitution $\ln\left(1+\frac{2}{\pi}u\right)=t$ and $1+\frac{2}{\pi}u=e^{t}$ but forgot to change du to dt. doing so yields $\frac{\pi^{2}}{4}\int_{ }^{ }\frac{e^{u}-e^{2u}}{u}$ which results in an Ei(u) function as the answer $\endgroup$
    – Mihindu
    Commented Sep 30, 2023 at 17:56
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    $\begingroup$ @Ronald F answer, according to the Ei(x) formula is $\frac{\pi^{2}}{4}\left(Ei\left(u\right)-Ei\left(2u\right)\right)+c$ which when resubstituted gives $\frac{\pi^{2}}{4}\left(Ei\left(\ln\left(1+\frac{1}{\pi}\cos^{-1}\left(x\right)\right)\right)-Ei\left(2\ln\left(1+\frac{2}{\pi}\cos^{-1}\left(x\right)\right)\right)\right)+c$. Still not elementary functions, and Im not sure why the definite integral causes contradictions $\endgroup$
    – Mihindu
    Commented Sep 30, 2023 at 18:00
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    $\begingroup$ @Ronald F but by integrating by parts recursively, you get the sequence $\frac{e^{2u}}{2u}+\frac{e^{2u}}{2u^{2}}+2\left(\frac{e^{2u}}{2u^{3}}+3\left(\frac{e^{2u}}{2u^{4}}+4...\right)\right)$ which simplifies to $\frac{e^{2u}}{2u}\left(1+\frac{1}{u}+\frac{2}{u^{2}}+\frac{6}{u^{3}}+\frac{24}{u^{4}}...\right)$ which is = $\frac{e^{2u}}{2u}\sum_{n=0}^{\infty}\left(\frac{r!}{u^{r}}\right)$. doing the same for e^u/u gives $\frac{e^{u}}{u}\sum_{n=0}^{\infty}\left(\frac{r!}{u^{r}}\right)$. $\endgroup$
    – Mihindu
    Commented Sep 30, 2023 at 18:10
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    $\begingroup$ If $\frac{e^{u}}{u}\sum_{n=0}^{\infty}\left(\frac{r!}{u^{r}}\right)$ = f(u), then $\frac{e^{2u}}{2u}\sum_{n=0}^{\infty}\left(\frac{r!}{u^{r}}\right)$ = f(2u), which lines up with Wolfram alphas answer of Ei(u) and Ei(2u), but the Wikipedia page for the integral doesn't have the exact summation I have. Also, would the value diverge as factorials grow faster than exponential/power functions? So how would a definite integral be found if it always diverges? $\endgroup$
    – Mihindu
    Commented Sep 30, 2023 at 18:13
  • $\begingroup$ Also a note, my substitutions were $x\ =\ \cos\left(\theta\right)\ and\ \ln\left(1+\frac{2}{\pi}\theta\right)=u$. I changed it in my first comment in the begining, changing theta to u and u to t to match your working, but in the last line, the integral in u would be in t according to my substitution, and the following working of Ei(u) and other summations is the integral after the second substitution, which would be when you substituted t. sorry its a bit confusing $\endgroup$
    – Mihindu
    Commented Sep 30, 2023 at 18:16

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