Integrating, the $\int_0^\infty \frac{\text{ d}x}{x^2\ln x+1}$ I am trying to find a closed form for
$$\int_0^\infty \frac{\text{ d}x}{x^2\ln x+1}$$
My usual tactic of partial differentiation under the integral (Mellin) does not work here due to that pesky +1 
 A: There is no close form for this integral. However, Mathematica says that it does converge to this result:
$$2.0313(...)$$
This integral is however problematic in many ways. First of all we cannot evaluate it with standard methods of integration. We may try for example to split it into
$$\left(\int_0^1 + \int_1^{+\infty}\right) \frac{\text{d}x}{x^2\ln(x) + 1}$$
The first one is evaluable by using the Geometric Series:
$$\int_0^1 \text{d}x\left(1 - x^2\ln(x) + x^4\ln^2(x) - \cdots \right)$$
Make the substitution
$$\ln(x) = y ~~~~~~~ \text{d}x = e^y\ \text{d}y$$
$$\int_{-\infty}^0\ \text{d}y\ e^y\left(1 - e^{2y}y + e^{4y}y^2 + \cdots \right)$$
$$\int_0^{+\infty}\ \text{d}y\ e^{-y}\left(1 + y e^{-2y} + y^2e^{-4y} + \cdots \right)$$
Those integrals are evaluable in the same way with the help of Gamma Function, and you will get
$$\Gamma(1) + \frac{1}{3^2}\Gamma(2) + \frac{1}{5^3}\Gamma(3) + \cdots$$
We can easily guess the inherent series that would arise with more and more terms, namely:
$$\sum_{k = 0}^{+\infty} \frac{1}{(2k+1)^{k+1}}\Gamma(k+1)$$
The very first terms are
$$1 + \frac{1}{9} + \frac{2}{125} + \frac{6}{2401} + \frac{24}{59049} + \frac{120}{1771561} + \cdots \approx 1.1301$$
I have no reason to think that the above series converges to that number, but surely the series does converge. However, a numerical integration about the first integral with Mathematica really gives us that number as a result. So my method is correct, for this one.
This is what you can do for the first integral.
Now, the second one is really evil
$$\int_1^{+\infty} \frac{1}{x^2\ln(x) + 1}\ \text{d}x$$
Indeed we cannot use a Geometric Series. An attempt to do that would show a divergence.
According to Mathematica, the final result is the one above I wrote, and this means that
$$\int_1^{+\infty} \frac{1}{x^2\ln(x) + 1}\ \text{d}x = 2.0313 - 1.1301 = 0.9012$$
Actually, currently I have no idea about how to proceed for this integral. It's surely something beautiful but I have to think more.
This is all, for the moment.
