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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2The integral seems indeed to be non-solvable (in the usual sense of a closed form), per WA. – Ben Grossmann Jan 11 '17 at 19:57
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1Presumably, a proof of non-integrability would involve Liouville's theorem, but I don't know much about differential Galois theory – Ben Grossmann Jan 11 '17 at 20:01
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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. – Amin235 Jan 11 '17 at 20:13
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@Armin235 That integral also seems to not be solvable. 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. – Ben Grossmann Jan 11 '17 at 20:16
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2You might want to check here: http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – Simply Beautiful Art Jan 11 '17 at 20:16
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@SimpleArt Thanks for introducing useful link. – Amin235 Jan 11 '17 at 20:19
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2If 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. – Pentapolis Jan 12 '17 at 06:14
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@Pentapolis Is it possible to address a reference for your hint, thanks. – Amin235 Jan 12 '17 at 09:35
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@Amin235 That integral requires elliptic integrals – Тyma Gaidash Dec 15 '22 at 13:05
2 Answers
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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.
Simply Beautiful Art
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You changed the form of integral that theorem of Liouville cab be applied. Am I right? – Amin235 Jan 11 '17 at 20:31
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@Amin235 Huh? I just used u-substitution and definitions of special functions. – Simply Beautiful Art Jan 11 '17 at 20:32
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@Amin235 Oh god, do you mean this? Um, hm... yeah...nope. Didn't use that at all >.> – Simply Beautiful Art Jan 11 '17 at 20:35
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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."
– Amin235 Jan 11 '17 at 20:37 -
@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. – Simply Beautiful Art Jan 11 '17 at 20:38
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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 $x\geq 1$.
Claude Leibovici
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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. – Amin235 Jan 12 '17 at 09:28