Integrating, the $\int_0^\infty \frac{\text{ d}x}{x^2\ln x+1}$

Question

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

Try a change of variable : z=x+1 , dz=dx , x²=(z-1)² = z²-2z+1 — Arif Burhan, Mar 17 '16 at 00:55
The numerical value is more or less
$$2.0313$$ so no elegant form will arise (namely, in terms of math famous constants). I'll provide to write down a solution tomorrow because I'm dying of sleep. — Enrico M., Mar 17 '16 at 02:20
@1over137 I'm curios how you can make the conclusion that this has no elegant form based on a numerical solution? — mike van der naald, Mar 17 '16 at 02:27
@mikevandernaald I didn't base my opinion on a numerical solution, but on MY solution. — Enrico M., Mar 17 '16 at 13:32

score 1 · Answer 1 · answered Mar 18 '16 at 19:35

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.

u can use $x\rightarrow 1/x$ to bring ur second integral into a form that can be expanded in terms of a power series — tired, Mar 19 '16 at 12:28
@tired
With $x\to \frac{1}{x}$ I'll get the integral

$$\int_0^1 \frac{\text{d}x}{\ln\left(\frac{1}{x}\right) + 1}$$

Which is a well know integral which gives

$$-e\ \text{Ei}(-1)$$

whose value is 0.5963(...) and it's not enough :/ But nice idea! — Enrico M., Mar 19 '16 at 13:20
@1over137, you made a mistake - with the substitution $x \to 1/x$ you get: $$\int_0^1 \frac{dx}{x^2-\ln(x)} \approx 0.9012$$ — Yuriy S, Mar 24 '16 at 08:04
The problem is - you separated the integral at $x=1$, but you should have done it at $x=\exp( \frac{W(2)}{2} )=1.5316\dots$, which is the solution for $x^2 \ln(x)=1$ — Yuriy S, Mar 26 '16 at 18:46

Integrating, the $\int_0^\infty \frac{\text{ d}x}{x^2\ln x+1}$

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