How to show that $ \int_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} dt=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$

Question

I've already known that $$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=\frac{1}{4\pi\sqrt {7}}\Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7})$$

To get this answer, I let $\mathrm{d}u=\mathrm{d}\sqrt{t}$, then got

$$ \int \limits_0^\infty \frac{1}{\sqrt {t(t^2+21t+112)}} \mathrm{d}t=2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u$$

$$2\int \limits_0^\infty \frac{1}{\sqrt {u^4+21u^2+112}} \mathrm{d}u=2\int \limits_0^\infty \frac{1}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \mathrm{d}u$$

and it was pretty like Incomplete elliptic integral of the first kind.

But, how to carry on?

Answer 1

As noted, this integral is reducible to an elliptic integral and the form of the proposed solution may suggest that its modulus could be a singular value. Indeed, from the form \begin{align} I&=2\int_0^\infty \frac{\mathrm{d}u}{\sqrt {u^4+21u^2+112}} \\ &=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {u^2+\frac{21}{2}+\sqrt{\frac{7}{4}}i}{\sqrt {u^2+\frac{21}{2}-\sqrt{\frac{7}{4}}i}}} \\ &=2\int_{0}^\infty \frac{\mathrm{d}u}{\sqrt {\left( u^2+u_+^2\right)\left( u^2+u_-^2 \right)}} \end{align} The opposite of the roots of the polynomial under the square root ($u_{\pm}^2=(21\pm\sqrt{7})/2$) are complex conjugate, as the coefficients of the polynomial are real. In the following, we first apply a Landen's transformation where these roots are replaced by their geometric and arithmetic means which are real.

Defining $p=\sqrt{u_+u_-}$ and $q=(u_++u_-)/2$, we find from the coefficients of the polynomial \begin{align} p^2&=\sqrt{112}=4\sqrt{7}\\ q^2&=\frac{1}{4}\left( u_+^2+u_-^2+2u_+u_- \right)\\ &=\frac{1}{4}\left(21+2\sqrt{112} \right)=\frac{1}{4}\left(21+8\sqrt{7} \right) \end{align} With $u=x+\sqrt{x^2+p^2}$, remarking that \begin{equation} \sqrt {\left( u^2+u_+^2\right)\left(u^2+u_-^2 \right)}=2u\sqrt{x^2+q^2} \end{equation} and that \begin{equation} du=\frac{u}{\sqrt{x^2+p^2}} \end{equation} The integral becomes \begin{equation} I=\int_{-\infty}^\infty\frac{dx}{\sqrt{\left( x^2+p^2 \right)\left( x^2+q^2 \right)}} \end{equation} Now, changing $x=q\tan\theta$ and taking into account the parity of the function, one obtains an integral representation of the complete elliptic integral of the first kind: \begin{align} I&=2\int_0^{\pi/2} \frac{d\theta}{\sqrt{p^2\cos^2\theta+q^2\sin^2\theta}}\\ &=\frac{2}{p}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{p^2-q^2}{p^2}\sin^2\theta}}\\ &=\frac{2}{p}\mathbf{K}\left( \sqrt{1-\frac{q^2}{p^2}} \right)%\\ %&=\frac{2}{p}K'\left( \frac{q}{p} \right) \end{align} Now, $\sqrt{1-\frac{q^2}{p^2}} =\frac{\sqrt{2}}{8}\left( 3-\sqrt{7} \right)=\lambda^*(7)$, where $\lambda^*(r)$ gives a singular value of the elliptic modulus $k_r$ for which $\mathbf{K}\left( \sqrt{1-k_r^2} \right)=\sqrt{r}\mathbf{K}(k_r)$ see here, here and here. Then, \begin{align} I&=\frac{2}{p}\mathbf{K}\left( k_7 \right)\\ &=\frac{2}{2.7^{1/4}}\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi 7^{1/4}}\\ &=\frac{\Gamma\left( \frac{1}{7} \right)\Gamma\left( \frac{2}{7} \right)\Gamma\left( \frac{4}{7} \right)}{4\pi \sqrt{7}} \end{align} as expected.

Answer 2

Perform the transformation: \begin{align*} \left\{ \begin{split} x(s)&=4\cdot\tfrac{(1+s^{7})^{1/7}+s(1+s^{7})^{5/7}-s^{5}}{s^{2}(1+s^{7})^{2/7}}-4\\ y(s)&=4+12\cdot\tfrac{(1+s^{7})^{1/7}+s(1+s^{7})^{5/7}-s^{5}}{s^{2}(1+s^{7})^{2/7}}+8\cdot\tfrac{(1+s^{7})^{5/7}-s^{4}-s^{5}(1+s^{7})^{4/7}}{s^{3}(1+s^{7})^{3/7}}\\ \end{split}\right. \end{align*}

After extensive algebraic simplification, we obtain: \begin{align*} \left|\dfrac{{\mathrm{d}}x}{y}\right|&=\dfrac{{\mathrm{d}}x}{\sqrt{x(x^2+21x+112)}}\\ &=\dfrac{(1+s^{7})^{3/7}+s^{3}(1+s^{7})^{1/7}-s}{(1+s^{7})^{6/7}}{\mathrm{d}}s\qquad(s>0) \end{align*} Note that $x(s)$ is a 2-covering from the positive real numbers to the interval $[0, +\infty)$, and it attains its minimum value of 0 at some point $\alpha$ on the positive real line. Therefore: \begin{align*} I_{s}&=\int_{0}^{+\infty}\frac{(1+s^{7})^{3/7}+s^{3}(1+s^{7})^{1/7}-s}{(1+s^{7})^{6/7}}{\mathrm{d}}s\\ &= \int_{0}^{\alpha}\frac{(1+s^{7})^{3/7}+s^{3}(1+s^{7})^{1/7}-s}{(1+s^{7})^{6/7}}{\mathrm{d}}s\\ &\qquad+\int_{\alpha}^{+\infty}\frac{(1+s^{7})^{3/7}+s^{3}(1+s^{7})^{1/7}-s}{(1+s^{7})^{6/7}}{\mathrm{d}}s \end{align*} $$=2\int_{0}^{+\infty}\frac{{\mathrm{d}}x}{\sqrt{x(x^{2}+21x+112)}}$$

The integral $I_s$ can be evaluated using the Euler integral formula, yielding an expression involving the Gamma function. \begin{aligned} \int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}} &=\frac{1}{m} \int_{0}^{1} y^{\left(\alpha-\frac{n+1}{m}\right)-1}(1-y)^{\frac{n+1}{m}-1} d y \\ &=\frac{1}{m} B\left(\alpha-\frac{n+1}{m}, \frac{n+1}{m}\right) \\ &=\frac{1}{m} \frac{\Gamma\left(\alpha-\frac{n+1}{m}\right) \Gamma\left(\frac{n+1}{m}\right)}{\Gamma(\alpha)} \end{aligned}

The parameterization of the elliptic curve $E: y^2 = x(x^2 + 21x + 112)$ using the Fermat septic $\lambda^7 + \mu^7 + \nu^7 = 0$ can be achieved through a series of transformations and mappings, as described in Gilles Lachaud's paper Ramanujan Modular Forms and the Klein Quartic (link).

1. Klein Quartic and the Elliptic Curve $E_1$

The paper provides a mapping from the Klein quartic $U^3V + V^3W + W^3U = 0$ to an elliptic curve $E_1$ with complex multiplication by $\mathbb{Z}[\sqrt{-7}]$. The elliptic curve $E_1$ is given by: $$ Y^2 + 3XY + Y = X^3 - 2X - 3. $$ This elliptic curve $E_1$ is isomorphic to the given elliptic curve $E: y^2 = x(x^2 + 21x + 112)$.

2. Mapping from Fermat Septic to Klein Quartic

The Fermat septic is defined by $\lambda^7 + \mu^7 + \nu^7 = 0$. There exists a rational map from the Fermat septic to the Klein quartic: $$ (U, V, W) = (\lambda^3 \nu, \lambda \mu^3, \mu \nu^3). $$ In affine form, this map can be written as: $$ (U, V, W) = \left( -\frac{\lambda^3}{\mu \nu^2}, -\frac{\lambda \mu^2}{\nu^3}, -1 \right). $$

3. Irrational Parametrization

To simplify the parameterization, we use the following irrational parametrization: \begin{align*} \left\{ \begin{split} \lambda &= (1 + s^7)^{1/7}, \\ \mu &= -s, \\ \nu &= -1. \end{split} \right. \end{align*} Substituting these into the affine map, we obtain: \begin{align*} \left\{ \begin{split} U &= \tfrac{(1 + s^7)^{3/7}}{s}, \\ V &= {\scriptsize{s}^2 (1 + s^7)^{1/7}}, \\ W &= -1. \end{split} \right. \end{align*}

4. Mapping from Klein Quartic to $E_1$

The paper provides explicit formulas for $X$ and $Y$ in terms of $U, V, W$: \begin{align*} \left\{ \begin{split} X &= -\left(\tfrac{V}{U} + \tfrac{U}{W} + \tfrac{W}{V} \right), \\ Y &= \tfrac{U}{V} + \tfrac{W}{U} + \tfrac{V}{W}. \end{split} \right. \end{align*} Substituting the expressions for $U, V, W$ from the irrational parametrization, we obtain: \begin{align*} \left\{ \begin{split} X&=\tfrac{(1+s^{7})^{1/7}+s(1+s^{7})^{5/7}-s^{5}}{s^{2}(1+s^{7})^{2/7}},\\ Y&=\tfrac{(1+s^{7})^{5/7}-s^{4}-s^{5}(1+s^{7})^{4/7}}{s^{3}(1+s^{7})^{3/7}}. \end{split} \right. \end{align*}

5. Isomorphism Between $E_1$ and $E$

The elliptic curves $E_1$ and $E$ are isomorphic, with the following explicit maps: \begin{cases} E_1 \to E: & (X, Y) \mapsto (x, y) = (4(X - 1), 4(1 + 3X + 2Y)), \\ E \to E_1: & (x, y) \mapsto (X, Y) = \left( \frac{1}{4}(x - 3), \frac{1}{8}(5 - 3x + y) \right). \end{cases}

6. Final Parameterization of $E$

Combining the above transformations, we obtain the parameterization of $E$ in terms of $s$: \begin{align*} \left\{ \begin{split} x&=4(X-1), \\ y&=4(1+3X+2Y), \\ X&=\tfrac{(1+s^{7})^{1/7}+s(1+s^{7})^{5/7}-s^{5}}{s^{2}(1+s^{7})^{2/7}},\\ Y&=\tfrac{(1+s^{7})^{5/7}-s^{4}-s^{5}(1+s^{7})^{4/7}}{s^{3}(1+s^{7})^{3/7}}. \end{split} \right. \end{align*}

Y^2 + 3*X*Y + Y - X^3 + 2*X + 3 /. {X -> -(U/W + V/U + W/V), 
    Y -> U/V + V/W + W/U} /. {U -> (1 + s^7)^(3/7)/s, 
   V -> s^2 (1 + s^7)^(1/7), W -> -1} // Factor
y^2 - x*(x^2 + 21*x + 112) /. {x -> (X - r)/u^2, 
    y -> (Y - s (X - r) - t)/u^3} /. 
  Thread[{u, r, s, t} -> {1/2, 1, -3/2, -2}] // Factor