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Is there a closed form for the following integral?

$$\int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx$$

It is approximately equal to $-0.48878092308456029189008$.

Mathematica is unable to find a closed form for this integral.

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    $\begingroup$ In case you need only the closed form: $$\frac{\Gamma \left(\frac{1}{8}\right) \Gamma \left(\frac{3}{8}\right)\sqrt{1+\sqrt{2}} }{8 \sqrt[4]{2} \sqrt{\pi }} \left\{\frac{\sqrt{2}-1}{4} \log \left(2+2 \sqrt{2}\right)-\frac{2-\sqrt{2}}{4} \pi \right\}$$ I got this using formula $4.242$ of Gradshteyn and Ryzhik's tables. $\endgroup$ Mar 13, 2014 at 12:32

1 Answer 1

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We let $$I(a,b)=\int^1_0\frac{x^{2a}}{\sqrt{1-x^2}\sqrt{x^2+b}}dx$$ so that OP's integral is equal to $\frac12\frac{d}{da}I(0,2+2\sqrt2)$.

According to Mathematica, we have $$I(a,b)=\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}{}_2F_1(\frac12,a+\frac12;a+1\mid-1/b)\\ =\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}(1+\frac1b)^{-1/2}{}_2F_1(\frac12,\frac12;a+1\mid\frac{1}{b+1}) =\frac{\sqrt\pi}{2\sqrt{b+1}}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}{}_2F_1(\frac12,\frac12;a+1\mid\frac{1}{b+1}).$$

I'm currently looking for suitable transforms to relate $I(a,b)$ to $I(-a,b)$. It looks like DLMF15.8.4 will be a good candidate.

Edit: Using DLMF15.8.4, we have $$\sin(a\pi)\sqrt{b}I(a,b)=\cos(a\pi)I(-a,1/b)-b^aI(a,1/b).$$ Taking derivative with respect to $a$: $$\sin(a\pi)\sqrt{b}I_a(a,b)+\pi\cos(a\pi)\sqrt{b}I(a,b)=-\cos(a\pi)I_a(-a,1/b)-\pi\sin(a\pi)I(-a,1/b)-b^aI_a(a,1/b)+b^a\log bI(a,1/b).$$

Let $a=0$:

$$\pi\sqrt{b}I(0,b)=-I_a(0,1/b)-I_a(0,1/b)-\log bI(0,1/b).$$

Noting $$I(0,b)=\frac{K\left(\sqrt{\frac{1}{1+b}}\right)}{\sqrt{1+b}},$$ we have $$I_a(0,1/b)=-\frac12(\pi\sqrt{b}I(0,b)+\log bI(0,1/b))\\ =-\frac12\sqrt{\frac{b}{1+b}}\left(\pi K\left(\sqrt{\frac{1}{1+b}}\right)+\log b K\left(\sqrt{\frac{b}{1+b}}\right)\right).$$

Thus, we can get OP's integral by taking $b=1/(2+2\sqrt2)=\frac{\sqrt2-1}{2}$:

$$I=\frac12I_a(0,2+2\sqrt2)=-\frac{\sqrt2-1}{4}\left(\pi K\left(\sqrt{2\sqrt2-2}\right)+\log\left(\frac{\sqrt2-1}{2}\right) K\left(\sqrt2-1\right)\right)\\ =-\frac{\sqrt2-1}{4}\left(\pi K'\left(\sqrt2-1\right)-\log\left(2\sqrt2+2\right) K\left(\sqrt2-1\right)\right).$$

Now note that $\sqrt2-1=k_2$ is the second Elliptic Integral Singular Value. Using the value of $K(k_2)$ there, we conclude that $$I=-\frac{\sqrt2-1}{4}\left(\pi K'\left(\sqrt2-1\right)-\log\left(2\sqrt2+2\right) K\left(\sqrt2-1\right)\right)\\ =-\frac{\sqrt2-1}{4}K(k_2)\left(\pi \frac{K'(k_2)}{K(k_2)}-\log\left(2\sqrt2+2\right)\right)\\ =-\frac{\sqrt2-1}{4}\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)\sqrt{\sqrt2+1}}{2^{13/4}\sqrt{\pi}}\left(\sqrt{2}\pi-\log\left(2\sqrt2+2\right)\right)\\ =-\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)\sqrt{\sqrt2-1}}{2^{21/4}\sqrt{\pi}}\left(\sqrt{2}\pi-\log\left(2\sqrt2+2\right)\right).$$

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