# Proving $\int_{0}^{1}{1\over \sqrt[4]{\ln \left({1\over x}\right)+\ln^2\left({1\over x}\right)}}\cdot{\mathrm dx\over x}=\cdots$

Consider this integral $(1)$

$$\int_{0}^{1}{1\over \sqrt[4]{\ln \left({1\over x}\right)+\ln^2\left({1\over x}\right)}}\cdot{\mathrm dx\over x}=-\Gamma\left(-{2\over 4}\right)\cdot{\Gamma\left({3\over 4}\right)\over \Gamma\left({1\over 4}\right)}\tag1$$

How can one prove $(1)$?

An attempt:

Rewrite $(1)$ as

$$\int_{0}^{1}(-\ln x+\ln^2(x))^{-1/4}\cdot{\mathrm dx\over x}\tag2$$

$u=\ln x \implies x\mathrm du =\mathrm dx$ then $(2)$ becomes

$$\int_{0}^{\infty}(u^2-u)^{-1/4}\mathrm du\tag3$$

May be we can split it into partial decomposition of fraction

$${1\over u^{1/4}(u-1)^{1/4}}={A\over u^{1/4}}+{B\over (u-1)^{1/4}}$$

Then $(3)$ becomes

$$\int_{0}^{\infty}{\color{red}{A\over u^{1/4}}}+{B\over (u-1)^{1/4}}\mathrm du\tag4$$

But the red part diverges, how else can we tackle $(1)?$

• you messed up your substitution somehow..the correct intermediate result is $$\int_0^{\infty}\frac{du}{(u(1+u))^{1/4}}$$ which should be simple enough to integrate for you – tired Mar 22 '17 at 7:36
• In addition to the comment by @tired, it might be more convenient to do the substitution $u=-\ln x$. – mickep Mar 22 '17 at 7:44
• You correctly wrote May be we can split it into partial decomposition of fractions but we cannot. What would be $A,B$ except functions of $u$ ? – Claude Leibovici Mar 22 '17 at 8:34

Let's consider $I=\displaystyle \int_{0}^1 {1\over \sqrt[4]{\ln \left({1\over x}\right)+\ln^2\left({1\over x}\right)}}\cdot{\mathrm dx\over x}$.
By the change of variable $$u=-\ln x, \quad du= - \frac{dx}x,$$ one gets $$I=\displaystyle \int_0^{\infty} {1\over \sqrt[4]{u+u^2}}\cdot{\mathrm du}$$ but, as $u \to \infty$, one has $${1\over \sqrt[4]{u+u^2}} \sim{1\over \sqrt[4]{u^2}}=\frac1{u^{1/2}}$$ and the latter integrand gives a divergent integral near $\infty$.