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$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).
The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx = \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

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    $\begingroup$ The answer to question $2$ is Stirling's approximation evaluated at $n=e$ (i.e. the approximation for $e!$). It seems related, but I'm not sure exactly how. $\endgroup$ Jan 7 '21 at 0:12
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First, notice that $\log(x)+1=\log(ex)$. With this, first substitute $u=ex$, then $k=\log u$. Your integral is now $$e^{1/2}\int_0^\infty k^{1/2} \,e^{-k/2}\,dk.$$ This is a well known integral that can be evaluated in terms of the Gamma function: $$e^{1/2}\,\Gamma(3/2)\,2^{3/2}=\sqrt{2\pi e}.$$ I used $\Gamma(3/2)=\sqrt{\pi}/2$. To learn more about the gamma function, see here. You can think about it as an analytic continution of the factorial $n!$ for $n\in\mathbb{C}$ instead of just $n\in\mathbb{N}$.

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    $\begingroup$ That's for (2). (1) doesn't seem to have a closed form. $\endgroup$ Jan 7 '21 at 0:28
  • $\begingroup$ I'm currently working on and thinking about (1). I'll let you know if I find anything interesting. $\endgroup$
    – Zachary
    Jan 7 '21 at 3:47
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    $\begingroup$ Now this is very nice. $\endgroup$
    – K.defaoite
    Jan 7 '21 at 4:01
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Concerning $(3)$ $$\int_{\frac 1e}^{\infty} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx $$ it is normal to get a complex result.

Around $x=\frac 1e$ we have $$\sqrt{\frac{\log{(x)}}{x^3 + 1}}=i\, \sum_{n=0}^\infty a_n \left(x-\frac{1}{e}\right)^n$$ and the first coefficients are $$a_0=\frac{ e^{3/2}}{\left(1+e^3\right)^{1/2}}\qquad a_1 =-\frac{e^{5/2} \left(4+e^3\right)}{2 \left(1+e^3\right)^{3/2}}\qquad a_2 =\frac{e^{7/2} \left(22-4 e^3+e^6\right)}{8 \left(1+e^3\right)^{5/2}}$$ So $$\int_{\frac 1e}^{1} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx = i\, \sum_{n=0} ^\infty \frac {(e - 1)^{n + 1} } {(n+1) e^{n+1} }a_n$$ using the given terms, this leads to $$\int_{\frac 1e}^{1} \sqrt{\frac{\log{x}}{x^3 + 1}}\,dx \sim 0.373129 \,i$$

In French, we have an expression which says "This, Sir, is the cause of your daughter's being dumb". In French, this simply means "This explains that !"

Edit

Consider $$\sqrt{\frac{\log{(x)}}{x^3 + a}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^{-3 n-\frac{3}{2}} \sqrt{\log (x)} \,a^n$$ This gives $$\int\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=$$ $$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} \left(\frac{\sqrt{2 \pi } \text{erf}\left(\sqrt{3 n+\frac{1}{2}} \sqrt{\log (x)}\right)}{(6 n+1)^{3/2}}-\frac{2 x^{-3 n-\frac{1}{2}} \sqrt{\log (x)}}{6 n+1}\right)\,a^n$$

$$\int_1^\infty\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=\sqrt{2\pi}\sum_{n=0}^\infty \frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}\,a^n$$

For $a=1$, computing the partial sums $$S_p=\sqrt{2\pi}\sum_{n=0}^p\frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}$$ $$\left( \begin{array}{cc} p & S_p \\ 10 & 2.45333 \\ 20 & 2.45303 \\ 30 & 2.45297 \\ 40 & 2.45294 \\ 50 & 2.45293 \\ 60 & 2.45293 \\ 70 & 2.45292 \end{array} \right)$$ which is the result given by numerical integration (this number is not recognized by inverse symbolic calculators).

This sum is alternating and we have $$\frac{S_{p+2}}{S_p}=1-\frac{4}{p}+\frac{51}{4 p^2}+O\left(\frac{1}{p^3}\right)$$ which explains the rather slow convergence.

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  • $\begingroup$ Am I that dumb? :-( $\endgroup$
    – user757932
    Jan 18 '21 at 18:11
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    $\begingroup$ @AFVAC. Sorry if you took that for you ! This French expression means "This explains that !". I was explaining why we got a complex number; that'all. Sorry for the confusion ! I am still working the problem. Cheers :-) $\endgroup$ Jan 19 '21 at 4:52
  • $\begingroup$ Thanks for your time to doing this problem sir, thank you! $\endgroup$
    – user757932
    Jan 19 '21 at 5:45
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    $\begingroup$ @AFVAC. I am almost finishing ! Good news (I hope). $\endgroup$ Jan 19 '21 at 5:47
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I don't think integral (1) has a closed form. It can be represented by a series, though:

$$ \sum_{k=0}^\infty \frac{2 \sqrt{\pi}\, {\mathrm e}^{3 k +\frac{1}{2}} \left(-1\right)^{k} \left(2 k \right)! 4^{-k}}{\left(6 k +1\right) \sqrt{12 k +2}\, k !^{2}}$$

This comes from replacing $x^3+1$ by $x^3+a$, expanding the integrand in a series in powers of $a$, integrating term-by-term, and then substituting $a=1$.

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    $\begingroup$ Are you sure you can interpret this as a convergent series in the classical sense? The summand asymptotically behaves as $e^{3k}/k^2$, meaning the series only converges pointwise for $|a| \le e^{-3}$, and so certainly not at $a=1$. I think this may have to do with the fact that swiching sum and integral destroys the convergence of the initial power series as you are integrating on a non-compact interval. Now you may be able to interpret it as an asymptotic series, however... $\endgroup$
    – Zachary
    Jan 7 '21 at 4:15
  • $\begingroup$ Yes, that's a good point. Summability methods must be used. $\endgroup$ Jan 7 '21 at 4:38
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    $\begingroup$ You also might want to mention in your answer that the series doesn't converge in the usual sense of a limit of partial sums, since that is what most people assume when they see one. The series isn't even asymptotic since there is no $a$ dependence, though it might be possible to use resummation techniques or try to analytically continue the original power series beyond its radius of convergence in order to make better sense of it. $\endgroup$
    – Zachary
    Jan 7 '21 at 6:10
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$$I=\int_{1/e}^\infty\sqrt{\frac{\ln x+1}{x^3+1}}dx$$ notice that: $\ln x+1=\ln x+\ln e=\ln ex$ if we try $u=ex\Rightarrow dx=du/e$ and so: $$I=\frac1e\int_1^\infty\sqrt{\frac{\ln u}{(u/e)^3+1}}du=\sqrt{e}\int_1^\infty\sqrt{\frac{\ln u}{u^3+e^3}}du$$ now this is quite a tough integral


$$J=\int_{1/e}^\infty\sqrt{\frac{\ln x+1}{x^3}}dx=\sqrt{e}\int_1^\infty u^{-3/2}\sqrt{\ln u}\,du$$ if we now let $v=\ln u\Rightarrow du=e^vdv$ so: $$J=\sqrt{e}\int_0^\infty e^{-1/2v}v^{1/2}dv$$ now let $w=\frac12v,dv=2dw$ $$J=2\sqrt{2e}\int_0^\infty e^{-w}w^{1/2}dw=2\sqrt{2e}\Gamma\left(\frac32\right)=2\sqrt{2e}\times\frac{\sqrt{\pi}}{2}=\sqrt{2e\pi}$$

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For the first integral, write $$I=\int_{1/e}^\infty\sqrt{\frac{\log x+1}{x^3+1}}\,dx=\int_0^\infty\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=I_1+I_2$$ where $$I_1=\int_0^1\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\int_0^1u^{1/2}e^{(3n+1)(u-1)}\,du$$ and $$I_2=\int_1^\infty\frac{u^{1/2}e^{u-1}}{(1+e^{3(u-1)})^{1/2}}\,du=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\int_1^\infty u^{1/2}e^{-(3n+1/2)(u-1)}\,du$$ so $$I=\sum_{n\ge0}\frac{(2n-1)!!(-1)^n}{2^nn!}\left[\frac1{(3n+1)^{3/2}e^{3n+1}}\int_0^1v^{1/2}e^v\,dv+\frac{e^{3n+1/2}}{(3n+1/2)^{3/2}}\int_1^\infty v^{1/2}e^{-v}\,dv\right]$$ where the integrals are incomplete gamma functions.

For the third integral, another way to see that it is not real is to note that $$\Im\int_{1/e}^\infty\sqrt{\frac{\log x}{x^3+1}}\,dx=\Im\int_{1/e}^1\sqrt{\frac{\log x}{x^3+1}}\,dx=\Im\int_1^e\sqrt{\frac{-\log u}{u^4+u}}\,du\ne0$$ as $\log u/(u^4+u)$ is positive on $(1,\infty)$.

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