7

I have it on good authority that the following monstrosity

$$I=\int_0^1 \sin\left(\sqrt{\frac{1-x}{x}}-\sqrt{\frac{x}{1-x}}\right)\frac{dx}{x\left(3x^2 - 3x +1\right)}$$

is not only convergent but has an analytic closed form. After spending a long time struggling with it I was not able to convert it into a familiar form, mainly because of the nested function within the $\sin(\cdot)$ term which I am unable to get rid of.

The essential singularities of the integrand about $x=0$ and $x=1$ make it difficult even to approximate the integral to any meaningful precision, so I am unable even to formulate a conjecture as to the closed form.

I am hoping that some kind soul will be able to rid me of this suspense and solve the devil-incarnate that is this integral.

Ng Chung Tak
  • 18,990
user1892304
  • 2,808

1 Answers1

8

$$\int_{0}^{1}\sin\left(\frac{1-2x}{\sqrt{x(1-x)}}\right)\frac{dx}{x(3x^2-3x+1)}\stackrel{x\mapsto\frac{1+z}{2}}{=}-4\int_{-1}^{1}\sin\left(\frac{2z}{\sqrt{1-z^2}}\right)\frac{dz}{(1+z)(1+3z^2)}$$ Through the substitution $z=\frac{t}{\sqrt{4+t^2}}$ the last integral becomes

$$ \int_{-\infty}^{+\infty}\sin(t)\frac{t-\sqrt{4+t^2}}{1+t^2}\,dt =\int_{-\infty}^{+\infty}\frac{t\sin(t)}{1+t^2}\,dt = \color{red}{\frac{\pi}{e}}$$ due to parity and the residue theorem.

Jack D'Aurizio
  • 353,855
  • 1
    Could you elaborate on how your note about the symmetry of the $\sin(\cdots)$ part of the integral led you to this solution? – Chris Apr 16 '17 at 19:02
  • @Chris: such symmetry leads to the RHS of the first line that is way more promising than the starting point. After that it is enough to perform a change of variable, in order to turn $\frac{2z}{\sqrt{1-z^2}}$ into $t$, to brilliantly solve the mistery. – Jack D'Aurizio Apr 16 '17 at 19:04
  • 2
    @JackD'Aurizio I'm quite amazed by the efficiency of your solution. What you did in three substitutions would have taken me ten. Nice work. – user1892304 Apr 16 '17 at 19:45