The integral has a closed-form solution $\int_0^1\frac{1}{\sqrt{\frac1x+x}+\sqrt{\frac1x-x}} \, \mathrm{d}x$?

Question

Here is my question:

Does this integral has a closed-form solution ? \begin{equation} \int_0^1\frac{1}{\sqrt{\frac1x+x}+\sqrt{\frac1x-x}} \, \mathrm{d}x\approx0.350927 \end{equation}

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I tried to transform it. $$\int_0^1\frac{1}{\sqrt{\frac1x+x}+\sqrt{\frac1x-x}} \,\mathrm{d}x=\int_0^1\frac{\sqrt x}{\sqrt{1+x^2}+\sqrt{1-x^2}} \,\mathrm{d}x=?$$ But I still couldn’t obtain a result. I suspect this is related to elliptic integrals, and the answer might be expressible in terms of special values of the Gamma function.

Answer 1

Multiplied by the conjugate, the integrand becomes $$f(x)=\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{2 x^{3/2}}$$ $$f(x)=\frac{1}{4 \sqrt{\pi }}\sum_{n=0}^\infty\left(1-(-1)^n\right)\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n+1)}\,x^{2 n-\frac{3}{2}}$$

$$\int f(x)\,dx=\frac{1}{2 \sqrt{\pi }}\sum_{n=0}^\infty\left(1-(-1)^n\right)\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{(4n-1)\,\Gamma (n+1)}\,x^{2 n-\frac{1}{2}}$$ and the antiderivative is given in terms of two Gaussian hypergeometric functions $$\frac 1{\sqrt x}\Bigg(\, _2F_1\left(-\frac{1}{2},-\frac{1}{4};\frac{3}{4 };x^2\right)-\, _2F_1\left(-\frac{1}{2},-\frac{1}{4};\frac{3}{4 };-x^2\right) \Bigg)$$ which make the definite integral $$ \frac{2 \left(\sqrt{2}-1\right) \pi ^{3/2}} {\Gamma\left(\frac{1}{4}\right)^2} $$

Edit

Starting from @Svyatoslav's approach $$I(t)=\frac 1{\sqrt 2} \int \left(\sqrt{\cosh (t)}-\sqrt{\sinh(t)}\right) \,dt$$ using elliptic integrals $$I(t)=-\sqrt{2} \left(\sqrt{i} E\left(\left.\frac{1}{4} (\pi -2 i t)\right|2\right)+i E\left(\left.\frac{i t}{2}\right|2\right)\right)$$

$$I(0)=-\frac{(1+i) \left(2 E\left(\frac{1}{2}\right)-K\left(\frac {1}{2}\right)\right)}{\sqrt{2}}=-\frac{(1+i)\, \Gamma\left(\frac{3}{4}\right)^2}{\sqrt{2 \pi }}$$ and when $t\to \infty$ $$I(t)\to -\frac{\left(1+(-1)^{3/4}\right) \Gamma \left(\frac{3}{4}\right)^2}{\sqrt{\pi }}$$ and then the result $$\frac{\sqrt{2}-1}{\sqrt{\pi }}\, \Gamma \left(\frac{3}{4}\right)^2$$

Answer 2

An interesting integral :) $$I=\int_0^1\frac{\sqrt x}{\sqrt{1+x^2}+\sqrt{1-x^2}} dx\overset {x=e^{-t}}{=}\int_0^\infty\frac{e^{-t}}{\sqrt{e^t+e^{-t}}+\sqrt{e^t-e^{-t}}}dt$$ $$=\frac1{\sqrt2}\int_0^\infty\frac{\cosh t-\sinh t}{\sqrt{\cosh t}+\sqrt{\sinh t}}dt=\frac1{\sqrt2}\int_0^\infty\left(\sqrt{\cosh t}-\sqrt{\sinh t}\right)dt$$ Now, let's take $\alpha\in(0,1)$ and consider $$I\alpha=\frac1{\sqrt2}\int_0^\infty\left(\cosh^{-\alpha} t-\sinh^{-\alpha} t\right)dt=J_1+J_2$$ where $$J_1=\frac{2^\alpha}{\sqrt2}\int_0^\infty\frac{dt}{(e^t+e^{-t})^\alpha}=2^{\alpha-\frac12}\int_0^\infty\frac{e^{-\alpha t}}{(1+e^{-2t})^\alpha}dt$$ $$\overset{x=e^{-2t}}{=}2^{\alpha-\frac32}\int_0^1\frac{x^{\frac \alpha2-1}}{(1+x)^\alpha}dx$$ Making the substitution $t=\frac1x$ $$J_1=2^{\alpha-\frac32}\int_1^\infty\frac{t^{\frac \alpha2-1}}{(1+t)^\alpha}dt\,\Rightarrow J_1=\frac12\,2^{\alpha-\frac32}\int_0^\infty\frac{t^{\frac \alpha2-1}}{(1+t)^\alpha}dt$$ Making the substitution $x=\frac1{1+t}$ $$J_1=2^{\alpha-\frac52}\int_0^1(1-x)^{\frac\alpha2-1}x^{\frac\alpha2-1}dx=2^{\alpha-\frac52}\frac{\Gamma^2\big(\frac\alpha2\big)}{\Gamma(\alpha)}\tag{1}$$ Using the same approach, $$J_2=-2^{\alpha-\frac32}\int_0^1\frac{x^{\frac \alpha2-1}}{(1-x)^\alpha}dx=-2^{\alpha-\frac32}\frac{\Gamma(1-\alpha)\Gamma\big(\frac\alpha2\big)}{\Gamma\big(1-\frac\alpha2\big)}\tag{2}$$ Hence, $$I_\alpha=J_1+J_2=2^{\alpha-\frac32}\left(\frac12\frac{\Gamma^2\big(\frac\alpha2\big)}{\Gamma(\alpha)}-\frac{\Gamma(1-\alpha)\Gamma\big(\frac\alpha2\big)}{\Gamma\big(1-\frac\alpha2\big)}\right)\tag{3}$$ We have got an analytical (with respect to $\alpha$) function $I\alpha$. Now, taking $\alpha=-\frac12$, we get the value of the desired integral: $$I=I_{-\frac12}=2^{-2}\left(\frac12\frac{\Gamma^2\big(-\frac14\big)}{\Gamma\big(-\frac12\big)}-\frac{\Gamma\big(\frac32\big)\Gamma\big(-\frac14\big)}{\Gamma\big(\frac54\big)}\right)$$ Using the properties of Gamma-function, $$-\frac14\Gamma\Big(-\frac14\Big)=\Gamma\Big(\frac34\Big);\quad-\frac12\Gamma\Big(-\frac12\Big)=\sqrt\pi;\quad\Gamma\Big(\frac14\Big)\Gamma\Big(\frac34\Big)=\pi\sqrt2$$ we get the answer in the form, presented by @Idividedbyzero in the comment above: $$I=\frac{\sqrt2-1}{\sqrt\pi}\Gamma^2\Big(\frac34\Big)$$

Answer 3

$$I=\int_{0}^{1}\frac{dx}{\sqrt{\frac{1}{x}+x}+\sqrt{\frac{1}{x}-x}} \overset{x=\frac{1-t}{1+t}}=\int_{0}^{1}\frac{\left(\sqrt{2t^2+2}-2\sqrt{t}\right)dt}{(1-t)^{3/2}(1+t)^{3/2}}\\\overset{(1)}=\int_{0}^{1}\frac{\sqrt{x}dx}{\sqrt{1-x^2}}-\sqrt{2}\int_{0}^{1}\frac{x^2dx}{\sqrt{1-x^4}}\\=\left(1-\frac{\sqrt{2}}{2}\right)\int_{0}^{1}\frac{\sqrt{x}dx}{\sqrt{1-x^2}}\overset{x=\sqrt{t}}=\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right)\int_{0}^{1}\frac{dt}{t^{1/4}(1-t)^{1/2}}\\\overset{(2)}=\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right)B(\frac{3}{4},\frac{1}{2})=\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right)\frac{\Gamma({3/4})\Gamma({1/2})}{\Gamma({5/4})}\overset{(3)}=\left(\frac{1}{2}-\frac{\sqrt{2}}{4}\right)\frac{4\sqrt{2}\pi^{3/2}}{\Gamma^2({1/4})}.$$

In $(1)$ we have integrated by parts ($u=\sqrt{2t^2+2}-2\sqrt{t},\hspace{.2cm} dv=1/(1-t^2)^{3/2}$).

In $(2)$ we have used Euler's Beta function: $$B(a,b)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma({a})\Gamma({b})}{\Gamma({a+b})}.$$

In $(3)$ we have used Euler's reflection formula: $$\Gamma({z})\Gamma({1-z})=\frac{\pi}{\sin({\pi z})}.$$