Estimating $\int_0^1 a^x\ \frac{x(1-x)}{\sin(\pi x)} dx$

calculusdefinite integralsimproper-integralsintegrationreal-analysis

I doubt a closed form exists, so I am trying to approximate the integral:
$$I(a)=\int_0^1 a^x\ \frac{x(1-x)}{\sin(\pi x)} dx$$
I am therefore looking for a function $F(a)$ that provides
$$F(a)\simeq I(a)$$ But I want this to be true for every value of $a>0$, and so something like a truncated Taylor series of $I$ would be useless, since for large $a$ it would require adding more and more terms.

Up to this day I found only two promising approximations of $I(a)$:

$\Large{1)}$

Using the inequality (which is a quite good approximation for small values of $x,y$)
$$\frac{x-y}{\log x-\log y}<\left(\frac{x^{\frac13}+y^{\frac23}}{2} \right)^3 $$
I got

$$F_1(a)=\frac{1+a}{8\pi}+\frac{a^{\frac13}+a^{\frac23}}{6\sqrt3}$$

with a percentage error of roughly $1\%$ for small $a$, growing larger as $a$ grows.

$\Large{2)}$

Using the famous ancient approximation for $\sin x$:
$$\sin x \simeq \frac{16(\pi-x)x}{5\pi^2-4(\pi-x)x}$$
I got

$$F_2(a)=\frac{a-1}{2\log^3a}-\frac{a+1}{4\log^2a}+\frac{5(a-1)}{16\log a}$$

and this one is tremendously precise, even for very large values of $a$.

However, I am still on the lookout for other, even better, approximations for $I$. These two are the only ones I was able to get. Regarding the first one, I tried even sharper bounds known for the logarithmic mean, but then I wasn't able to integrate. The generalized mean of exponent $\frac13$ was the best I could get. I like the result because it features only polynomials, no exponentials or logarithms and trig functions. But, since the two means start to differ very fast past around $10$, the result is good only for small values of $a$.

If someone has an idea to get another approximation I'd be happy to hear it.

Best Answer

The OP asked a quick question about this question where @user170231 answered expanding a special case of @Mariusz Iwaniuk’s hypergeometric function. If we expand the general case, then the polygamma function appears:

$$\,_4\text F_3\left(2,b,b,b;b+1,b+1,b+1,b+1;1\right)=\sum_{n=0}^\infty\frac{(2)_n(b)_n^3}{(b+1)_n^3n!}=b^3\sum_{n=0}^\infty\frac{n+1}{(n+b)^3}=\frac{b^3}2\left(2\psi^{(1)}(b)+(b-1)\psi^{(2)}(b)\right)$$

Therefore:

$$\boxed{\int_0^1 a^xx(1-x)\csc(\pi x)dx=\frac{i(a-1)}{2\pi^2}\psi^{(1)}\left(\frac{i\ln(a)}{2\pi}+\frac12\right)-\frac{(a+1)}{4\pi^3}\psi^{(2)}\left(\frac{i\ln(a)}{2\pi}+\frac12\right)}$$

shown here:

Related Solutions

Calculus – Proving $\int_0^\infty\left({_2F_1}\left(\frac{1}{6},\frac{1}{2};\frac{1}{3};-x\right)\right)^{12}dx=\frac{80663}{153090}$

Consider the hypergeometric equation with parameters $(a,b,c)=\left(\frac16,\frac12,\frac13\right)$, and build from its two canonical solutions near $z=0$ the vector $$\vec{y}(z)=\left(\begin{array}{c} y_1 \\ y_2 \end{array}\right)=\left(\begin{array}{c} _2F_1(a,b;c;z) \\ z^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;z) \end{array}\right).\tag{1}$$ This is a single-valued vector function on $\mathbb{C}\backslash\{(-\infty,0]\cup[1,\infty)\}$. Its analytic continuation along a closed loop $\gamma$ gives rise to monodromy representation of $\pi_1(\mathbb{C}\backslash\{0,1\})$: $$ \gamma\mapsto M_{[\gamma]},\qquad y(\gamma z)=M_{[\gamma]}y(z).$$ The monodromy group $G\subset GL(2,\mathbb{C})$ of the hypergeometric equation is generated by two matrices corresponding to simple loops around $0$ and $1$. In the case we are interested in these matrices are explicitly given by $$M_0=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{-2\pi i /3}\end{array}\right),\qquad M_1=C\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{2\pi i /3}\end{array}\right)C^{-1},\tag{2}$$ where the connection matrix $C=\left(\begin{array}{cc} 1 & 2^{\frac43} \\ -2^{\frac83} & 8\end{array}\right)$. If $G$ is finite, then $\vec{y}(z)$ has a finite number of branches, and moreover (Schwarz 1872), is algebraic.

It is not difficult to check that the monodromy group $G$ generated by $M_0$, $M_1$ from (2) is indeed finite. In particular, note that $$M_0^3=M_1^3=I,\qquad M_1^2=-M_0M_1M_0, $$ $$M_1M_0M_1=-M_0^2,\qquad M_1M_0^2M_1=M_0^2M_1M_0^2.$$ It turns out that $G$ has order $24$ and is isomorphic to the binary tetrahedral group: $$G\cong 2T=\langle s,t\,|\,(st)^2=s^3=t^3\rangle, $$ where the generators can be identified as $s=M_0M_1M_0M_1M_0$, $t=M_1M_0M_1M_0M_1$.

Corollary: The hypergeometric functions in (1) are algebraic.

Algebraic solutions of the hypergeometric equations are classified by the so-called Schwarz table, and have been studied by many mathematicians, see e.g. the bibliography in this paper by R.Vidunas. Their explicit construction is somewhat involved but relatively straightforward - at least when the corresponding algebraic curve has genus $0$ (the genus can be determined independently from the Riemann-Hurwitz formula).

In our case the task simplifies even more as our parameter values can be obtained from the genus $0$ tetrahedral formula (2.4) of the above mentioned paper by a combination of a linear trasformation (sending $\frac56$ to $\frac43-\frac56=\frac12$) and differentiation (transforming $\frac43$ into $\frac13$). The result is $$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}.$$

Corollary: The antiderivative $\displaystyle\int \mathcal{R}\left(x,y(x)\right)dx$, where $y(x)={}_2F_1\left(\frac16,\frac12;\frac13;-x\right)$ and $\mathcal{R}(x,y)$ is rational in both arguments, can be expressed in terms of elementary functions.

Example: The transformation $r\mapsto x(r)=\frac{r(r+2)^3}{(r+1)(1-r)^3}$ bijectively maps $(0,1)$ to $(0,\infty)$, and therefore the initial integral becomes \begin{align}\mathcal{I}&=\int_0^1 \left(\frac{\sqrt{1-r^2}}{2r+1}\right)^{12}\left(\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)'dr=\\&= 2\int_0^1 \frac{(1+r)^4(1-r)^2(r+2)^2}{(2r+1)^{10}} dr=\\&=\frac{80\,663}{153\,090}. \end{align}

Calculus – Prove ${\large\int}_0^1\frac{\ln(1+8x)}{x^{2/3}\,(1-x)^{2/3}\,(1+8x)^{1/3}}dx=\frac{\ln3}{\pi\sqrt3}\Gamma^3\!\left(\tfrac13\right)$

Define $\mathcal{I}$ to be the value of the definite integral,

$$\mathcal{I}:=\int_{0}^{1}\frac{\ln{\left(1+8x\right)}}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\,\mathrm{d}x\approx3.8817.$$

Problem. Prove that the following conjectured value for the definite integral $\mathcal{I}$ is correct: $$\mathcal{I}=\int_{0}^{1}\frac{\ln{\left(1+8x\right)}}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\,\mathrm{d}x\stackrel{?}{=}\frac{\ln{(3)}}{\sqrt{3}\,\pi}\left[\Gamma{\left(\small{\frac13}\right)}\right]^3.\tag{1}$$

Elimination of logarithmic factor from the integrand:

Suppose we have a substitution relation of the form $1+8x=\frac{k}{1+8t}$, with $k$ being some positive real constant greater than $1$. First of all, the symmetry of the relation with respect to the variables $x$ and $t$ implies that $t$ solved for as a function of $x$ will have the same functional form as $x$ solved for as a function of $t$:

$$1+8x=\frac{k}{1+8t}\implies t=\frac{k-(1+8x)}{8(1+8x)},~~x=\frac{k-(1+8t)}{8(1+8t)}.$$

Transforming the integral $\mathcal{I}$ via this substitution, we find:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}x^{-2/3}\,(1-x)^{-2/3}\,(1+8x)^{-1/3}\,\ln{(1+8x)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\ln{(1+8x)}}{\sqrt[3]{x^{2}\,(1-x)^{2}\,(1+8x)}}\,\mathrm{d}x\\ &=\int_{\frac{k-1}{8}}^{\frac{k-9}{72}}\sqrt[3]{\frac{2^{12}(1+8t)^{5}}{k(9-k+72t)^{2}(k-1-8t)^{2}}}\,\ln{\left(\frac{k}{1+8t}\right)}\cdot\frac{(-k)}{(1+8t)^2}\,\mathrm{d}t\\ &=\left(\frac{k}{9}\right)^{2/3}\int_{\frac{k-9}{72}}^{\frac{k-1}{8}}\frac{\ln{\left(\frac{k}{1+8t}\right)}}{\left(\frac{9-k}{72}+t\right)^{2/3}\left(\frac{k-1}{8}-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t,\\ \end{align}$$

which clearly suggests the choice $k=9$ as being the simplest, in which case:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln{\left(\frac{9}{1+8t}\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\ln{\left(9\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t-\int_{0}^{1}\frac{\ln{\left(1+8t\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t\\ &=2\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}-\mathcal{I}\\ \implies 2\mathcal{I}&=2\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\\ \implies \mathcal{I}&=\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\\ &=:\ln{(3)}\,\mathcal{J}, \end{align}$$

where in the last line we've simply introduced the symbol $\mathcal{J}$ to denote the last integral for convenience. It's approximate value is $\mathcal{J}\approx3.53328$.

Thus, to prove that the conjectured value $(1)$ is indeed correct, it suffices to prove the following equivalent conjecture:

$$\mathcal{J}:=\int_{0}^{1}\frac{\mathrm{d}x}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\stackrel{?}{=}\frac{1}{\sqrt{3}\,\pi}\left[\Gamma{\left(\small{\frac13}\right)}\right]^3.\tag{2}$$

Representation and manipulation of integral as a hypergeometric function:

Euler's integral representation for the Gauss hypergeometric function states that, for $\Re{\left(c\right)}>\Re{\left(b\right)}>0\land |\arg{\left(1-z\right)}|<\pi$, we have:

$$\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\mathrm{d}x=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}.$$

In particular, if we choose $z=-8$, $a=\frac13$, $c=\frac23$, and $b=\frac13$, then the conditions $\Re{\left(\frac23\right)}>\Re{\left(\frac13\right)}>0\land |\arg{\left(1-(-8)\right)}|=0<\pi$ are satisfied, and the integral on the left-hand-side of Euler's representation reduces to the integral $\mathcal{J}$. That is,:

$$\begin{align} \mathcal{J} &=\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac23}(1+8x)^{-\frac13}\mathrm{d}x\\ &=\operatorname{B}{\left(\frac13,\frac13\right)}\,{_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)}. \end{align}$$

Using the quadratic transformation,

$${_2F_1}{\left(a,b;2b;z\right)} = \left(\frac{1+\sqrt{1-z}}{2}\right)^{-2a} {_2F_1}{\left(a,a-b+\frac12;b+\frac12;\left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^2\right)},$$

with particular values $a=b=\frac13,z=-8$, we have the hypergeometric identity,

$${_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)} = 2^{-\frac23} {_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}.$$

Then, applying Euler's transformation,

$${_2F_1}{\left(a,b;c;z\right)}=(1-z)^{c-a-b}{_2F_1}{\left(c-a,c-b;c;z\right)},$$

we have,

$${_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}={_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}.$$

Now, Euler's integral representation for this hypergeometric function implies:

$${_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}=\frac{1}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x.$$

Hence,

$$\begin{align} \mathcal{J} &=\operatorname{B}{\left(\frac13,\frac13\right)}\,{_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}}\,{_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}}\,{_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}\,\operatorname{B}{\left(\frac13,\frac12\right)}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x.\\ \end{align}$$

The ratio of beta functions in the last line above simplifies considerably. The Legendre duplication formula for the gamma function states:

$$\Gamma{\left(2z\right)}=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma{\left(z\right)}\Gamma{\left(z+\frac12\right)}.$$

Letting $z=\frac13$ yields:

$$\Gamma{\left(\frac23\right)}=\frac{2^{-1/3}}{\sqrt{\pi}}\Gamma{\left(\frac13\right)}\Gamma{\left(\frac56\right)}.$$

Then, using the facts that $\operatorname{B}{\left(a,b\right)}=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(b\right)}}{\Gamma{\left(a+b\right)}}$ and $\Gamma{\left(\frac12\right)}=\sqrt{\pi}$, we can simplify the ratio of beta functions above considerably:

$$\begin{align} \frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{\operatorname{B}{\left(\frac13,\frac12\right)}} &=\frac{\left[\Gamma{\left(\frac13\right)}\right]^2\,\Gamma{\left(\frac56\right)}}{\Gamma{\left(\frac23\right)}\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(\frac13\right)}}\\ &=\frac{\Gamma{\left(\frac13\right)}\,\Gamma{\left(\frac56\right)}}{\sqrt{\pi}\,\Gamma{\left(\frac23\right)}}\\ &=\frac{\sqrt{\pi}\,\sqrt[3]{2}}{\sqrt{\pi}}\\ &=\sqrt[3]{2}. \end{align}$$

Thus,

$$\begin{align} \mathcal{J} &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}\,\operatorname{B}{\left(\frac13,\frac12\right)}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x\\ &=\frac{\sqrt[3]{2}}{2^{2/3}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x\\ &=\frac{1}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(1-\small{\frac14}x\right)}}\,\mathrm{d}x.\\ \end{align}$$

Reduction of integral to pseudo-elliptic integrals:

Substituting $x=t^3$ into the integral representation for $\mathcal{J}$, we reduce the problem to solving an integral whose integrand is the reciprocal square-root of a sixth-degree polynomial:

$$\begin{align} \mathcal{J} &=\frac{1}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(1-\small{\frac14}x\right)}}\,\mathrm{d}x\\ &=\frac{2}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(4-x\right)}}\,\mathrm{d}x\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\frac13\,x^{-\frac23}\,\mathrm{d}x}{\sqrt{\left(1-x\right)\left(4-x\right)}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^3\right)\left(4-t^3\right)}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4-5t^3+t^6}}.\\ \end{align}$$

Then, substituting $t=\sqrt[3]{2}\,u$ gives us a similar integrand except with a sixth-degree polynomial with symmetric coefficients:

$$\begin{align} \mathcal{J} &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4-5t^3+t^6}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\sqrt[3]{2}\,\mathrm{d}u}{\sqrt{4-10u^3+4u^6}}\\ &=3\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\mathrm{d}u}{\sqrt{1-\small{\frac52}u^3+u^6}}.\\ \end{align}$$

The form of the integral $\mathcal{J}$ found in the last line above is significant because it may be transformed into a pair of pseudo-elliptic integrals, which can subsequently be evaluated in terms of elementary functions and elliptic integrals. For more information on these types of integrals, see my question here.

Substituting $u=z-\sqrt{z^2-1}$ transforms the integral $\mathcal{J}$ into a sum of two pseudo-elliptic integrals:

$$\begin{align} \mathcal{J} &=3\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\mathrm{d}u}{\sqrt{1-\small{\frac52}u^3+u^6}}\\ &=\frac{3}{\sqrt{2}}\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(8z^3-6z-\frac52)}}+\frac{3}{\sqrt{2}}\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(8z^3-6z-\frac52)}}\\ &=3\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(16z^3-12z-5)}}+3\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(16z^3-12z-5)}}\\ &=\frac34\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}+\frac34\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}\\ &=:\frac34\,P_{+}+\frac34\,P_{-},\\ \end{align}$$

where in the last line we've introduced the auxiliary notation $P_{\pm}$ to denote the pair of integrals,

$$P_{\pm}:=\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z\pm1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}.$$

Evaluation of pseudo-elliptic integrals:

Again for convenience, we shall denote the lower integration limit in the integrals $P_{\pm}$ defined above by $2^{-2/3}+2^{-4/3}=:\alpha\approx1.026811$. In particular, this eases the factorization of the cubic polynomial in the denominators above:

$$\begin{align} 16x^3-12x-5 &=16\left(x-\alpha\right)\left(x^2+\alpha x+\alpha^2-\small{\frac34}\right)\\ &=16\left(x-\alpha\right)\left[\left(x+\frac{\alpha}{2}\right)^2+\frac34\left(\alpha^2-1\right)\right]. \end{align}$$

Then,

$$\begin{align} P_{\pm} &=\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x^3-\small{\frac34}x-\small{\frac{5}{16}})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x^3-\small{\frac34}x-\small{\frac{5}{16}})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x-\alpha)(x^2+\alpha x+\alpha^2-\small{\frac34})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)\left(x-\alpha\right)\left[\left(x+\frac{\alpha}{2}\right)^2+\small{\frac34}\left(\alpha^2-1\right)\right]}}.\\ \end{align}$$

EDIT: I'm beginning to fear this strategy of derivation may be ultimately fruitless. The good news is that there is a relatively compact way of expressing the two integrals above as incomplete elliptic integrals:

Assume $\alpha,\beta,u,m,n\in\mathbb{R}$ such that $\beta<\alpha<u$. Then proposition 3.145(1) of Gradshteyn's Table of Integrals, Series, and Products states: $$\begin{align} \small{\int_{\alpha}^{u}\frac{\mathrm{d}x}{\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left[\left(x-m\right)^2+n^2\right]}}} &=\small{\frac{1}{pq}F{\left(2\arctan{\sqrt{\frac{q\left(u-\alpha\right)}{p\left(u-\beta\right)}}},\frac12\sqrt{\frac{\left(p+q\right)^2+\left(\alpha-\beta\right)^2}{pq}}\right)},} \end{align}$$ where $(m-\alpha)^2+n^2=p^2$, and $(m-\beta)^2+n^2=q^2$.

The bad news is after plugging in all the appropriate values, the resulting arguments of the elliptic integrals are significantly more complex than I expected.

Best Answer

Related Solutions

Calculus – Proving $\int_0^\infty\left({_2F_1}\left(\frac{1}{6},\frac{1}{2};\frac{1}{3};-x\right)\right)^{12}dx=\frac{80663}{153090}$

Calculus – Prove ${\large\int}_0^1\frac{\ln(1+8x)}{x^{2/3}\,(1-x)^{2/3}\,(1+8x)^{1/3}}dx=\frac{\ln3}{\pi\sqrt3}\Gamma^3\!\left(\tfrac13\right)$

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