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Consider integral involving the modified Bessel functions of the first and second kinds (of order one) and sine function

$I(a, b, c) = \int_0^{\infty} \frac{\sin(ax)}{x} I_1(bx) K_1(cx) \mathrm{d}x$

For the case $b=c$, Mathematica gives the result

$I(a, b, b) = \frac{1}{m} E(m) + \left( 1-\frac{1}{m} \right) K(m), \qquad m = - \frac{4 b^2}{a^2}$

where $K(m)$ and $E(m)$ are the complete elliptic integral of the first and second kinds respectively (in Mathematica notation).

How can one derive the general result for $b \ne c$?

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  • 2
    $\begingroup$ It should be noted that this integral is not convergent if $b>c$, since then the exponential growth of $I_1(bx)$ dominates the exponential decay of $K_1(cx)$. On the other hand, the asymptotics of $I_1(bx)$ are oscillatory with exponential amplitudes; for that reason, the integral may be well-defined for $b>c$ in an appropriately regularized sense. $\endgroup$ – Semiclassical Sep 6 '15 at 1:18
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Given $a\in\mathbb{R}^{+}\land b\in\mathbb{R}^{+}\land c\in\mathbb{R}^{+}\land c\ge b$, define $I{\left(a,b,c\right)}$ via the definite integral

$$I{\left(a,b,c\right)}:=\int_{0}^{\infty}\frac{\sin{\left(ax\right)}\,I_{1}{\left(bx\right)}\,K_{1}{\left(cx\right)}}{x}\,\mathrm{d}x.$$


Recall that the modified Bessel functions of the first and second kind may be given, respectively by the integral representations

$$I_{\nu}{\left(z\right)}=\frac{z^{\nu}}{2^{\nu}\sqrt{\pi}\,\Gamma{\left(\nu+\frac12\right)}}\int_{-1}^{1}e^{-zt}\left(1-t^{2}\right)^{\nu-\frac12}\,\mathrm{d}t;~~~\small{\Re{\left(\nu\right)}>-\frac12},$$

and

$$K_{\nu}{\left(z\right)}=\frac{\sqrt{\pi}\,z^{\nu}}{2^{\nu}\,\Gamma{\left(\nu+\frac12\right)}}\int_{1}^{\infty}e^{-zt}\left(t^{2}-1\right)^{\nu-\frac12}\,\mathrm{d}t;~~~\small{\Re{\left(\nu\right)}>-\frac12\land\Re{\left(z\right)}>0}.$$

Expressing the Bessel function factors as integrals via the representations above allows us to rewrite the integral $I{\left(a,b,c\right)}$ as a triple integral with elementary integrand. By changing the order of integration, we can reduce this to a double integral with rational integrand.

Suppose $a>0\land c>b>0$. We find

$$\begin{align} I{\left(a,b,c\right)} &=\int_{0}^{\infty}\frac{\sin{\left(ax\right)}\,I_{1}{\left(bx\right)}\,K_{1}{\left(cx\right)}}{x}\,\mathrm{d}x\\ &=\int_{0}^{\infty}\mathrm{d}x\,\frac{\sin{\left(ax\right)}}{x}\cdot\frac{bx}{\pi}\int_{-1}^{1}\mathrm{d}t\,e^{-bxt}\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,cxe^{-cxu}\sqrt{u^{2}-1}\\ &=\frac{bc}{\pi}\int_{0}^{\infty}\mathrm{d}x\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,x\sin{\left(ax\right)}e^{-bxt}e^{-cxu}\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}\\ &=\frac{bc}{\pi}\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}x\,\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}\,x\sin{\left(ax\right)}e^{-\left(bt+cu\right)x}\\ &=\frac{bc}{\pi}\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}\int_{0}^{\infty}\mathrm{d}x\,x\sin{\left(ax\right)}e^{-\left(bt+cu\right)x}\\ &=\small{\frac{bc}{\pi\,a^{2}}\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}\int_{0}^{\infty}\mathrm{d}y\,y\sin{\left(y\right)}e^{-\left(\frac{bt+cu}{a}\right)y}};~~~\small{\left[ax=y\right]}\\ &=\frac{bc}{\pi\,a^{2}}\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}\frac{2\left(\frac{bt+cu}{a}\right)}{\left[\left(\frac{bt+cu}{a}\right)^{2}+1\right]^{2}}\\ &=\frac{2abc}{\pi}\int_{-1}^{1}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{\left(bt+cu\right)\sqrt{\left(1-t^{2}\right)\left(u^{2}-1\right)}}{\left[a^{2}+\left(bt+cu\right)^{2}\right]^{2}}\\ &=\frac{ab}{\pi}\int_{-1}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{2c\left(bt+cu\right)\sqrt{u^{2}-1}}{\left[a^{2}+\left(bt+cu\right)^{2}\right]^{2}}\\ &=\frac{ab}{\pi}\int_{-1}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\sqrt{u^{2}-1}};~~~\small{I.B.P.s}\\ &=\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\sqrt{u^{2}-1}}\\ &~~~~~+\frac{ab}{\pi}\int_{-1}^{0}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\sqrt{u^{2}-1}}\\ &=\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\sqrt{u^{2}-1}}\\ &~~~~~+\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u}{\left[a^{2}+\left(-bt+cu\right)^{2}\right]\sqrt{u^{2}-1}}\\ &=\small{\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u\left[a^{2}+\left(bt-cu\right)^{2}\right]}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\left[a^{2}+\left(bt-cu\right)^{2}\right]\sqrt{u^{2}-1}}}\\ &~~~~~\small{+\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{u\left[a^{2}+\left(bt+cu\right)^{2}\right]}{\left[a^{2}+\left(bt+cu\right)^{2}\right]\left[a^{2}+\left(bt-cu\right)^{2}\right]\sqrt{u^{2}-1}}}\\ &=\small{\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{1}^{\infty}\mathrm{d}u\,\frac{2u\left(c^{2}u^{2}+a^{2}+b^{2}t^{2}\right)}{\left[c^{4}u^{4}+2c^{2}\left(a^{2}-b^{2}t^{2}\right)u^{2}+\left(a^{2}+b^{2}t^{2}\right)^{2}\right]\sqrt{u^{2}-1}}}\\ &=\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\small{\times\int_{1}^{\infty}\mathrm{d}v\,\frac{\left(c^{2}v+a^{2}+b^{2}t^{2}\right)}{\left[c^{4}v^{2}+2c^{2}\left(a^{2}-b^{2}t^{2}\right)v+\left(a^{2}+b^{2}t^{2}\right)^{2}\right]\sqrt{v-1}}};~~~\small{\left[u^{2}=v\right]}\\ &=\frac{ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}w\,\frac{\left(c^{2}w+a^{2}+b^{2}t^{2}+c^{2}\right)}{\left[\left(c^{2}w+a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}\right]\sqrt{w}};~~~\small{\left[v-1=w\right]}\\ &=\frac{2ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}x\,\frac{c^{2}x^{2}+a^{2}+b^{2}t^{2}+c^{2}}{\left(c^{2}x^{2}+a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}};~~~\small{\left[\sqrt{w}=x\right]}.\\ \end{align}$$

Then,

$$\begin{align} I{\left(a,b,c\right)} &=\frac{2ab}{\pi}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{0}^{\infty}\mathrm{d}x\,\frac{c^{2}x^{2}+a^{2}+b^{2}t^{2}+c^{2}}{\left(c^{2}x^{2}+a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}\\ &=\frac{2ab}{\pi\,c}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{2}+a^{2}+b^{2}t^{2}+c^{2}}{\left(y^{2}+a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}};~~~\small{\left[cx=y\right]}\\ &=\frac{2ab}{\pi\,c}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{2}+a^{2}+b^{2}t^{2}+c^{2}}{y^{4}+2\left(a^{2}-b^{2}t^{2}+c^{2}\right)y^{2}+\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}\\ &=\frac{2ab}{\pi\,c}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{2}+r}{y^{4}+2py^{2}+p^{2}+q^{2}};~~~\small{\left[a^{2}-b^{2}t^{2}+c^{2}=:p\land2abt=:q\land a^{2}+b^{2}t^{2}+c^{2}=:r\right]}.\\ \end{align}$$

Now, given $\left(a,b,c,d\right)\in\mathbb{R}^{4}\land0<c\land0<d$, define the auxiliary function $J{\left(a,b;c,d\right)}$ to be the value of the integral

$$J{\left(a,b;c,d\right)}:=\int_{0}^{\infty}\mathrm{d}y\,\frac{ay^{2}+b}{y^{4}+2cy^{2}+c^{2}+d^{2}}.$$

A general expression for $J{\left(a,b;c,d\right)}$ is derived in the appendix below. Using the result obtained there, we arrive at

$$\begin{align} I{\left(a,b,c\right)} &=\frac{2ab}{\pi\,c}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{2}+r}{y^{4}+2py^{2}+p^{2}+q^{2}}\\ &=\frac{2ab}{\pi\,c}\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t^{2}}\,J{\left(1,r;p,q\right)}\\ &=\frac{ab}{\sqrt{2}\,c}\int_{0}^{1}\mathrm{d}t\,\frac{\left[r+\sqrt{p^{2}+q^{2}}\right]\sqrt{1-t^{2}}}{\sqrt{p^{2}+q^{2}}\sqrt{p+\sqrt{p^{2}+q^{2}}}}\\ &=\frac{ab}{\sqrt{2}\,c}\int_{0}^{1}\mathrm{d}t\,\\ &~~~~~\small{\times\frac{\left[a^{2}+b^{2}t^{2}+c^{2}+\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}\right]\sqrt{1-t^{2}}}{\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}\sqrt{a^{2}-b^{2}t^{2}+c^{2}+\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}}}}.\\ \end{align}$$

The cumbersome expression obtained for the integrand in the last line above is ultimately quite manageable through the magic of Euler substitutions. We obtain

$$\begin{align} I{\left(a,b,c\right)} &=\frac{ab}{\sqrt{2}\,c}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{1-t^{2}}}{\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}}\\ &~~~~~\times\frac{a^{2}+b^{2}t^{2}+c^{2}+\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}}{\sqrt{a^{2}-b^{2}t^{2}+c^{2}+\sqrt{\left(a^{2}-b^{2}t^{2}+c^{2}\right)^{2}+4a^{2}b^{2}t^{2}}}}\\ &=\frac{ab}{2\sqrt{2}\,c}\int_{0}^{1}\mathrm{d}u\,\frac{\sqrt{1-u}}{\sqrt{u}\sqrt{\left(a^{2}-b^{2}u+c^{2}\right)^{2}+4a^{2}b^{2}u}}\\ &~~~~~\times\frac{a^{2}+b^{2}u+c^{2}+\sqrt{\left(a^{2}-b^{2}u+c^{2}\right)^{2}+4a^{2}b^{2}u}}{\sqrt{a^{2}-b^{2}u+c^{2}+\sqrt{\left(a^{2}-b^{2}u+c^{2}\right)^{2}+4a^{2}b^{2}u}}};~~~\small{\left[t^{2}=u\right]}\\ &=-\frac{ab}{2\sqrt{2}\,c}\int_{\sqrt{\left(a^{2}-b^{2}+c^{2}\right)^{2}+4a^{2}b^{2}}-b^{2}}^{a^{2}+c^{2}}\mathrm{d}w\,\frac{\left(-1\right)\left[\left(w-a^{2}+c^{2}\right)^{2}+4a^{2}c^{2}\right]}{2b^{2}\left(w-a^{2}+c^{2}\right)^{2}}\\ &~~~~~\times\sqrt{\frac{\left(w+b^{2}\right)^{2}-\left[a^{2}+\left(b-c\right)^{2}\right]\left[a^{2}+\left(b+c\right)^{2}\right]}{\left(a^{2}+c^{2}-w\right)\left(a^{2}+c^{2}+w\right)}}\\ &~~~~~\times\frac{2\left(w-a^{2}+c^{2}\right)}{\left(w-a^{2}+c^{2}\right)^{2}+4a^{2}c^{2}}\\ &~~~~~\times\frac{2c^{2}\sqrt{w+a^{2}+c^{2}}}{\left(w-a^{2}+c^{2}\right)};~~~\small{\left[\sqrt{\left(a^{2}-b^{2}u+c^{2}\right)^{2}+4a^{2}b^{2}u}-b^{2}u=w\right]}\\ &=\frac{ac}{\sqrt{2}\,b}\int_{\sqrt{\left(a^{2}-b^{2}+c^{2}\right)^{2}+4a^{2}b^{2}}-b^{2}}^{a^{2}+c^{2}}\frac{\mathrm{d}w}{\left(w-a^{2}+c^{2}\right)^{2}}\\ &~~~~~\times\sqrt{\frac{\left(w+b^{2}\right)^{2}-\left[a^{2}+\left(b-c\right)^{2}\right]\left[a^{2}+\left(b+c\right)^{2}\right]}{\left(a^{2}+c^{2}-w\right)}}.\\ \end{align}$$

This last integral is in now in a form recognizable as an elliptic integral. A general evaluation of the requisite elliptic integral is presented in Appendix 2 below. Hence, we arrive at the expression

$$\begin{align} I{\left(a,b,c\right)} &=\frac{ac}{\sqrt{2}\,b}\,\mathcal{E}{\left(-b^{2}-\delta,-b^{2}+\delta,a^{2}+c^{2};a^{2}-c^{2}\right)},\\ \end{align}$$

where $\delta:=\sqrt{\left[a^{2}+\left(b-c\right)^{2}\right]\left[a^{2}+\left(b+c\right)^{2}\right]}$. Using the elliptic integral result obtained in the appendix, the evaluation of $I{\left(a,b,c\right)}$ is essentially complete, except for back-substituting the appropriate formulas for the parameters.


Appendix 1:

Given $\left(a,b,c,d\right)\in\mathbb{R}^{4}\land0<c\land0<d$, define the auxiliary parameter $\alpha:=\arctan{\left(\frac{d}{c}\right)}$. We obtain

$$\begin{align} J{\left(a,b;c,d\right)} &=\int_{0}^{\infty}\mathrm{d}y\,\frac{ay^{2}+b}{y^{4}+2cy^{2}+c^{2}+d^{2}}\\ &=\small{\sqrt[4]{c^{2}+d^{2}}\int_{0}^{\infty}\mathrm{d}t\,\frac{a\sqrt{c^{2}+d^{2}}\,t^{2}+b}{\left(c^{2}+d^{2}\right)t^{4}+2c\sqrt{c^{2}+d^{2}}\,t^{2}+c^{2}+d^{2}}};~~~\small{\left[y=\sqrt[4]{c^{2}+d^{2}}\,t\right]}\\ &=\frac{\sqrt[4]{c^{2}+d^{2}}}{\left(c^{2}+d^{2}\right)}\int_{0}^{\infty}\mathrm{d}t\,\frac{a\sqrt{c^{2}+d^{2}}\,t^{2}+b}{t^{4}+2\frac{c}{\sqrt{c^{2}+d^{2}}}\,t^{2}+1}\\ &=\sqrt{\frac{\cos^{3}{\left(\alpha\right)}}{c^{3}}}\int_{0}^{\infty}\mathrm{d}t\,\frac{act^{2}\sec{\left(\alpha\right)}+b}{t^{4}+2t^{2}\cos{\left(\alpha\right)}+1}\\ &=\frac{\sqrt{\cos{\left(\alpha\right)}}}{c\sqrt{c}}\int_{0}^{\infty}\mathrm{d}t\,\frac{act^{2}+b\left[1-2\sin^{2}{\left(\frac{\alpha}{2}\right)}\right]}{\left[t^{2}-2t\sin{\left(\frac{\alpha}{2}\right)}+1\right]\left[t^{2}+2t\sin{\left(\frac{\alpha}{2}\right)}+1\right]}\\ &=\frac{\sqrt{\cos{\left(\alpha\right)}}}{c\sqrt{c}}\int_{0}^{\infty}\mathrm{d}t\,\frac{act^{2}+b\left(1-2s^{2}\right)}{\left(t^{2}-2st+1\right)\left(t^{2}+2st+1\right)};~~~\small{\left[\sin{\left(\frac{\alpha}{2}\right)}=:s\right]}\\ &=\frac{\sqrt{\cos{\left(\alpha\right)}}}{c\sqrt{c}}\int_{0}^{\infty}\mathrm{d}t\,\bigg{[}\frac{2bs-4bs^{3}+\left(ac+2bs^{2}-b\right)t}{4s\left(t^{2}-2st+1\right)}\\ &~~~~~+\frac{2bs-4bs^{3}-\left(ac+2bs^{2}-b\right)t}{4s\left(t^{2}+2st+1\right)}\bigg{]};~~~\small{P.F.D.}\\ &=\small{\frac{\sqrt{\cos{\left(\alpha\right)}}}{c\sqrt{c}}\int_{0}^{\infty}\mathrm{d}t\,\bigg{[}\frac{2bs-4bs^{3}+\left(ac+2bs^{2}-b\right)s+\left(ac+2bs^{2}-b\right)\left(t-s\right)}{4s\left(t^{2}-2st+1\right)}}\\ &~~~~~+\frac{2bs-4bs^{3}+\left(ac+2bs^{2}-b\right)s-\left(ac+2bs^{2}-b\right)\left(t+s\right)}{4s\left(t^{2}+2st+1\right)}\bigg{]}\\ &=\frac{\sqrt{\cos{\left(\alpha\right)}}}{4c\sqrt{c}}\int_{0}^{\infty}\mathrm{d}t\,\bigg{[}\frac{ac+b\left(1-2s^{2}\right)}{\left(t^{2}-2st+1\right)}\\ &~~~~~+\frac{ac+b\left(1-2s^{2}\right)}{\left(t^{2}+2st+1\right)}\bigg{]}\\ &~~~~~+\frac{\sqrt{\cos{\left(\alpha\right)}}}{c\sqrt{c}}\cdot\frac{\left(ac+2bs^{2}-b\right)}{8s}\int_{0}^{\infty}\mathrm{d}t\,\bigg{[}\frac{2\left(t-s\right)}{\left(t^{2}-2st+1\right)}\\ &~~~~~-\frac{2\left(t+s\right)}{\left(t^{2}+2st+1\right)}\bigg{]}\\ &=\frac{\left[ac+b\cos{\left(\alpha\right)}\right]\sqrt{\cos{\left(\alpha\right)}}}{4c\sqrt{c}\sqrt{1-s^{2}}}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}t\,\bigg{[}\frac{\sqrt{1-s^{2}}}{t^{2}-2st+1}+\frac{\sqrt{1-s^{2}}}{t^{2}+2st+1}\bigg{]}\\ &=\frac{\left[ac+b\cos{\left(\alpha\right)}\right]\sqrt{\cos{\left(\alpha\right)}}}{4c\sqrt{c}\sqrt{1-s^{2}}}\\ &~~~~~\times\bigg{[}\arctan{\left(\frac{t-s}{\sqrt{1-s^{2}}}\right)}+\arctan{\left(\frac{t+s}{\sqrt{1-s^{2}}}\right)}\bigg{]}_{t=0}^{t=\infty}\\ &=\frac{\pi\left[ac+b\cos{\left(\alpha\right)}\right]\sqrt{\cos{\left(\alpha\right)}}}{4c\sqrt{c}\cos{\left(\frac{\alpha}{2}\right)}}\\ &=\frac{\pi\left[ac+b\cos{\left(\alpha\right)}\right]\sqrt{\cos{\left(\alpha\right)}}}{2\sqrt{2}\,c\sqrt{c}\sqrt{1+\cos{\left(\alpha\right)}}}\\ &=\frac{\pi\left[ac+\frac{bc}{\sqrt{c^{2}+d^{2}}}\right]}{2\sqrt{2}\,c\sqrt{c+\sqrt{c^{2}+d^{2}}}}\\ &=\frac{\pi\left[b+a\sqrt{c^{2}+d^{2}}\right]}{2\sqrt{2}\sqrt{c^{2}+d^{2}}\sqrt{c+\sqrt{c^{2}+d^{2}}}}.\\ \end{align}$$


Appendix 2:

Define the elliptic integral,

$$\mathcal{E}{\left(\alpha,\beta,\gamma;\lambda\right)}:=\int_{\beta}^{\gamma}\frac{1}{\left(x-\lambda\right)^{2}}\sqrt{\frac{\left(x-\alpha\right)\left(x-\beta\right)}{\left(\gamma-x\right)}}\,\mathrm{d}x;~~~\small{\alpha<\lambda<\beta<\gamma}.$$

Then, given $\alpha<\lambda<\beta<\gamma$ we have

$$\begin{align} \mathcal{E}{\left(\alpha,\beta,\gamma;\lambda\right)} &=\int_{\beta}^{\gamma}\frac{1}{\left(x-\lambda\right)^{2}}\sqrt{\frac{\left(x-\alpha\right)\left(x-\beta\right)}{\left(\gamma-x\right)}}\,\mathrm{d}x\\ &=\left[\left(\frac{1}{\gamma-\lambda}-\frac{1}{x-\lambda}\right)\sqrt{\frac{\left(x-\alpha\right)\left(x-\beta\right)}{\left(\gamma-x\right)}}\right]_{x=\beta}^{x=\gamma}\\ &~~~~~-\int_{\beta}^{\gamma}\left(\frac{1}{\gamma-\lambda}-\frac{1}{x-\lambda}\right)\frac{\alpha\beta-\alpha\gamma-\beta\gamma+2\gamma\,x-x^{2}}{2\left(\gamma-x\right)^{2}\sqrt{\frac{\left(x-\alpha\right)\left(x-\beta\right)}{\left(\gamma-x\right)}}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &=\small{0-0-\int_{\beta}^{\gamma}\frac{\left(-1\right)\left(\gamma-x\right)}{\left(\gamma-\lambda\right)\left(x-\lambda\right)}\cdot\frac{\alpha\beta-\alpha\gamma-\beta\gamma+2\gamma\,x-x^{2}}{2\left(\gamma-x\right)\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}\,\mathrm{d}x}\\ &=\frac{1}{2\left(\gamma-\lambda\right)}\int_{\beta}^{\gamma}\frac{\alpha\beta-\alpha\gamma-\beta\gamma+2\gamma\,x-x^{2}}{\left(x-\lambda\right)\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}\,\mathrm{d}x\\ &=\frac12\int_{\beta}^{\gamma}\frac{\mathrm{d}x}{\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}\\ &~~~~~+\frac{1}{2\left(\gamma-\lambda\right)}\int_{\beta}^{\gamma}\sqrt{\frac{\left(\gamma-x\right)}{\left(x-\alpha\right)\left(x-\beta\right)}}\,\mathrm{d}x\\ &~~~~~+\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\gamma-\lambda\right)}\int_{\beta}^{\gamma}\frac{\mathrm{d}x}{\left(x-\lambda\right)\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}.\\ \end{align}$$

Next, we transform the elliptic integrals via the linear fractional transformation

$$\frac{\left(\gamma-\alpha\right)\left(x-\beta\right)}{\left(\gamma-\beta\right)\left(x-\alpha\right)}=y.$$

We obtain

$$\begin{align} \mathcal{E}{\left(\alpha,\beta,\gamma;\lambda\right)} &=\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\gamma-\lambda\right)}\int_{\beta}^{\gamma}\frac{\mathrm{d}x}{\left(x-\lambda\right)\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}\\ &~~~~~+\frac{1}{2\left(\gamma-\lambda\right)}\int_{\beta}^{\gamma}\sqrt{\frac{\left(\gamma-x\right)}{\left(x-\alpha\right)\left(x-\beta\right)}}\,\mathrm{d}x\\ &~~~~~+\frac12\int_{\beta}^{\gamma}\frac{\mathrm{d}x}{\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left(\gamma-x\right)}}\\ &=\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\gamma-\lambda\right)}\int_{0}^{1}\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)}{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}\\ &~~~~~\times\frac{\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y}{\left(\gamma-\alpha\right)\left(\beta-\lambda\right)+\left(\gamma-\beta\right)\left(\lambda-\alpha\right)y}\\ &~~~~~\times\frac{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\,\mathrm{d}y\\ &~~~~~+\frac{1}{2\left(\gamma-\lambda\right)}\int_{0}^{1}\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)}{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}\cdot\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\left(1-y\right)}{\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y}\\ &~~~~~\times\frac{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\,\mathrm{d}y\\ &~~~~~+\frac12\int_{0}^{1}\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)}{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}\\ &~~~~~\times\frac{\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]^{2}}{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\,\mathrm{d}y;~~~\small{\left[\frac{\left(\gamma-\alpha\right)\left(x-\beta\right)}{\left(\gamma-\beta\right)\left(x-\alpha\right)}=y\right]}\\ &=\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\gamma-\lambda\right)}\int_{0}^{1}\frac{\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y}{\left(\gamma-\alpha\right)\left(\beta-\lambda\right)+\left(\gamma-\beta\right)\left(\lambda-\alpha\right)y}\\ &~~~~~\times\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\\ &~~~~~+\frac{1}{2\left(\gamma-\lambda\right)}\int_{0}^{1}\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)\left(1-y\right)}{\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y}\\ &~~~~~\times\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\\ &~~~~~+\frac12\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left[\left(\gamma-\alpha\right)-\left(\gamma-\beta\right)y\right]}}\\ &=\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{1-\mu\,y}{\left(\beta-\lambda\right)+\mu\left(\lambda-\alpha\right)y}\\ &~~~~~\times\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{\left(\gamma-\beta\right)}{2\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{1-y}{1-\mu\,y}\cdot\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{1}{2\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}};~~~\small{\left[\mu:=\frac{\gamma-\beta}{\gamma-\alpha}\right]},\\ \end{align}$$

and then,

$$\begin{align} \mathcal{E}{\left(\alpha,\beta,\gamma;\lambda\right)} &=\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\beta-\lambda\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\cdot\frac{\mu}{\nu}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}}{2\left(\beta-\lambda\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\cdot\frac{\nu-\mu}{\nu}\\ &~~~~~\times\int_{0}^{1}\frac{\mathrm{d}y}{\left(1-\nu\,y\right)\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{\left(\gamma-\beta\right)}{2\left(\gamma-\lambda\right)\mu\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~-\frac{\left(\gamma-\beta\right)\left(1-\mu\right)}{2\left(\gamma-\lambda\right)\mu\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)^{3}}}\\ &~~~~~+\frac{1}{2\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}};~~~\small{\left[\nu:=\frac{\mu\left(\alpha-\lambda\right)}{\left(\beta-\lambda\right)}\right]}\\ &=\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)}{2\left(\lambda-\alpha\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\beta-\alpha\right)\left(\gamma-\lambda\right)^{2}}{2\left(\lambda-\alpha\right)\left(\beta-\lambda\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\\ &~~~~~\times\int_{0}^{1}\frac{\mathrm{d}y}{\left(1-\nu\,y\right)\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~-\frac{\left(\beta-\alpha\right)}{2\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)^{3}}}\\ &=\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)}{2\left(\lambda-\alpha\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~+\frac{\left(\beta-\alpha\right)\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\beta-\alpha\right)\left(\gamma-\lambda\right)^{2}}{2\left(\lambda-\alpha\right)\left(\beta-\lambda\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\\ &~~~~~\times\int_{0}^{1}\frac{\mathrm{d}y}{\left(1-\nu\,y\right)\sqrt{y\left(1-y\right)\left(1-\mu\,y\right)}}\\ &~~~~~-\frac{\left(\gamma-\alpha\right)}{2\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\int_{0}^{1}\frac{\sqrt{1-\mu\,t}}{\sqrt{t\left(1-t\right)}}\,\mathrm{d}y;~~~\small{\left[\frac{1-y}{1-\mu\,y}=t\right]}\\ &=\frac{\pi\left(\beta-\alpha\right)\left(\gamma-\alpha\right)}{2\left(\lambda-\alpha\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\,{_2F_1}{\left(\frac12,\frac12;1;\mu\right)}\\ &~~~~~-\frac{\pi\left(\gamma-\alpha\right)}{2\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\,{_2F_1}{\left(-\frac12,\frac12;1;\mu\right)}\\ &~~~~~+\frac{\pi\left(\beta-\alpha\right)\left[\left(\gamma-\alpha\right)\left(\gamma-\beta\right)-\left(\gamma-\lambda\right)^{2}\right]}{2\left(\lambda-\alpha\right)\left(\beta-\lambda\right)\left(\gamma-\lambda\right)\sqrt{\gamma-\alpha}}\,F_{1}{\left(\frac12;1,\frac12;1;\nu,\mu\right)}.\blacksquare\\ \end{align}$$


$\endgroup$
  • 3
    $\begingroup$ Did you actually type all of this from scratch? $\endgroup$ – Mark Viola Dec 9 '16 at 4:46
  • 1
    $\begingroup$ @Dr.MV ...yes. I guess I kinda lost track of how long it was becoming. ^^; $\endgroup$ – David H Dec 9 '16 at 5:14
  • 1
    $\begingroup$ good lord...(+1) $\endgroup$ – tired Dec 15 '16 at 13:24

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