Some horrid integrals I am confused with $$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$
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 :-)
 A: 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.
A: 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 $a=1$.
A: 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 $n\in\mathbb{N}$.
A: $$I=\int_{1/e}^\infty\sqrt{\frac{\ln x+1}{x^3+1}}dx$$
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 $w=\frac12v,dv=2dw$
$$J=2\sqrt{2e}\int_0^\infty e^{-w}w^{1/2}dw=2\sqrt{2e}\Gamma\left(\frac32\right)=2\sqrt{2e}\times\frac{\sqrt{\pi}}{2}=\sqrt{2e\pi}$$
A: 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 $(1,\infty)$.
