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How to solve the following integral $$ \int\sqrt{ln(x)}\,dx $$ If the above integral is not solve-able, how to proof that the function $\sqrt{ln(x)}$ is not integrable.

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    The integral seems indeed to be non-solvable (in the usual sense of a closed form), per [WA](http://www.wolframalpha.com/input/?i=integral+sqrt(ln(x))+dx). – 2017-01-11
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    Presumably, a proof of non-integrability would involve [Liouville's theorem](https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)), but I don't know much about [differential Galois theory](https://en.wikipedia.org/wiki/Differential_Galois_theory) – 2017-01-11
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    @Omnomnomnom unfortunatlly I do not know about differential Galois theory. But I appreciate, if you proof this integral is not solve-able. You know, I have other simple integral like $$ \int\frac{dx}{\sqrt[3]{cos(x)}} $$ that I guess are not solve-able, but I do not know how to poof it. – 2017-01-11
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    @Armin235 That integral also seems to [not be solvable](http://www.wolframalpha.com/input/?i=integral+1%2F(cos(x))%5E(1%2F3)+dx). The point of my earlier comment is to tell you that "proving an integral is non-solvable" falls under the field of differential Galois theory. Although I know this much, I have never seen such a proof, nor do I know how to put one together. – 2017-01-11
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    You might want to check here: http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – 2017-01-11
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    @SimpleArt Thanks for introducing useful link. – 2017-01-11
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    If a function is non-solvable then its inverse function is non-solvable too, so $y=\sqrt{\ln x}$ is the inverse function of $x=e^{y^2}$ which is wellknown that it is non-solvable in elementray functions. – 2017-01-12
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    @Pentapolis Is it possible to address a reference for your hint, thanks. – 2017-01-12

2 Answers 2

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Let $x=e^{-u}$ and you get

$$\int_a^b\sqrt{\ln x}\ dx=-\int_{-\ln a}^{-\ln b}\sqrt{-u}e^{-u}\ du=i\left(\gamma(3/2,-\ln a)-\gamma(3/2,-\ln b)\right)$$

where $\gamma(a,x)$ is the lower incomplete gamma function.

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    You changed the form of integral that theorem of Liouville cab be applied. Am I right? – 2017-01-11
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    @Amin235 Huh? I just used u-substitution and definitions of special functions. – 2017-01-11
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    @Amin235 Oh god, do you mean [this](https://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis))? Um, hm... yeah...nope. Didn't use that at all >.> – 2017-01-11
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    in the link that you mentioned in the comment there is the following theorem. "It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ nonconstant, the antiderivative $$\int [f(x)\exp(g(x))] \, \mathrm dx$$ can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution to the differential equation: $$f = h' + hg$$ $\int e^{x^2} dx$ is another classic example of such a function with no elementary antiderivative." – 2017-01-11
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    @Amin235 Oh, I suppose I sneakily turned it into that form, but that was on accident. I was meaning to get to the gamma functions was all. – 2017-01-11
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    Ok, Thanks for response. – 2017-01-11
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    @Amin235 :P Ok. – 2017-01-11
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Considering $$I=\int\sqrt{\log( x)}\ dx$$ change variable $$\sqrt{\log( x)}=t\implies x=e^{t^2}\implies dx=2te^{t^2}\,dt$$ which makes $$I=\int 2t^2e^{t^2}\,dt$$ Now, integration by parts $$u=t\implies u'=dt$$ $$v'=2te^{t^2}\,dt\implies v=e^{t^2}$$ $$I=te^{t^2}-\int e^{t^2}\,dt=te^{t^2} -\frac{\sqrt{\pi }}{2} \text{erfi}(t)$$ where appears the imaginary error function.

Back to $x$, $$I=x \sqrt{\log (x)}-\frac{\sqrt{\pi }}{2} \text{erfi}\left(\sqrt{\log (x)}\right)$$ which is real if $x\geq 1$.

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    The user @omnomnomnom in the first comment mentioned this hint. I thank you for this hint that you said, but I think this is not an answer to my question. – 2017-01-12