Question Number 220730 by Nicholas666 last updated on 18/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}\left(\mathrm{1}\:−\:{x}\right)\left(\mathrm{1}\:+\:{kx}\right)}}\:{dx}\:,\:\left(−\mathrm{1}\:<\:{k}\:<\:\mathrm{1}\right)\:\:\: \\ $$$$ \\ $$
Answered by SdC355 last updated on 18/May/25
$$ \\ $$$$\int\:\:\:\frac{\mathrm{d}{z}}{\:\sqrt{{z}\left(\mathrm{1}−{z}\right)\left(\mathrm{1}+{kz}\right)}}=\frac{\mathrm{2}}{\:\sqrt{{k}}}\:\mathrm{F}\left(\boldsymbol{{i}}\centerdot\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{{z}−\mathrm{1}}}\right)\mid\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({z}\mid{a}\right)\:\mathrm{is}\:\mathrm{Elliptic}\:\mathrm{interal}\:\mathrm{first}\:\mathrm{kind}\:\mathrm{with} \\ $$$$\mathrm{parameter}\:\mathrm{a}=\kappa^{\mathrm{2}} \\ $$$$\mathrm{F}\left({z},{a}\right)=\int\:\:\frac{\mathrm{d}{z}}{\:\sqrt{\mathrm{1}−{a}\centerdot\mathrm{sin}^{\mathrm{2}} \left({z}\right)}} \\ $$
Commented by Frix last updated on 18/May/25
$$\mathrm{Show}\:\mathrm{the}\:\mathrm{path}. \\ $$$$\mathrm{We}'\mathrm{re}\:{not}\:\mathrm{here}\:\mathrm{to}\:\mathrm{simply}\:\mathrm{type}\:/\:\mathrm{copy}\:\mathrm{the} \\ $$$$\mathrm{answers}\:\mathrm{from}\:\mathrm{wolframalpha}.\mathrm{com} \\ $$
Commented by SdC355 last updated on 18/May/25
$$\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{easily}\:\mathrm{to}\:\mathrm{solve}\:''\int\:\:\frac{\mathrm{1}}{\:\sqrt{{z}\left({z}−\mathrm{1}\right)\left({kz}+\mathrm{1}\right)}}\:\mathrm{d}{z}'' \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{express}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{easy} \\ $$
Commented by Ghisom last updated on 18/May/25
$$\mathrm{if}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it}\:\Rightarrow\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{it} \\ $$
Answered by Ghisom last updated on 18/May/25
![Ω=∫_0 ^1 (dx/( x^(1/2) (1−x)^(1/2) (1+kx)^(1/2) ))= [t=(((1−x)^(1/2) )/( x^(1/2) )) → dx=−2x^(3/2) (1−x)^(1/2) dt] =2∫_0 ^∞ (dt/( (√(t^4 +(k+2)t^2 +(k+1)))))= [u=2arctan (t/((k+1)^(1/4) )) → dt=((t^2 +(k+1)^(1/2) )/(2(k+1)^(1/4) ))du] =(k+1)^(−1/4) ∫_0 ^π (du/( (√(1+(((1−(k+1)^(1/2) )^2 )/(4(k+1)^(1/2) ))sin^2 u))))= =2(k+1)^(−1/4) ∫_0 ^(π/2) (du/( (√(1−c^2 sin^2 u))))= =(k+1)^(−1/4) K (c^2 ) [elliptic K] K (c^2 ) =(2/(πagm (1, (√(1−c^2 ))))) c^2 =−(((1−(k+1)^(1/2) )^2 )/(4(k+1)^(1/2) )) ⇒ Ω=(2/(π(k+1)^(1/4) agm (1, ((1+(k+1)^(1/2) )/(2(k+1)^(1/4) ))))) agm (p, q) =lim_(n→∞) u_n =lim_(n→∞) v_n u_0 =p∧v_0 =q u_(n+1) =((u_n +v_n )/2)∧v_(n+1) =(u_n v_n )^(1/2)](https://www.tinkutara.com/question/Q220783.png)
$$\Omega=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\:{x}^{\mathrm{1}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{kx}\right)^{\mathrm{1}/\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} }{\:{x}^{\mathrm{1}/\mathrm{2}} }\:\:\rightarrow\:{dx}=−\mathrm{2}{x}^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{1}/\mathrm{2}} {dt}\right] \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\:\sqrt{{t}^{\mathrm{4}} +\left({k}+\mathrm{2}\right){t}^{\mathrm{2}} +\left({k}+\mathrm{1}\right)}}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{2arctan}\:\frac{{t}}{\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }\:\rightarrow\:{dt}=\frac{{t}^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} }{\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }{du}\right] \\ $$$$=\left({k}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{4}} \underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{du}}{\:\sqrt{\mathrm{1}+\frac{\left(\mathrm{1}−\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:{u}}}= \\ $$$$=\mathrm{2}\left({k}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{4}} \underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{du}}{\:\sqrt{\mathrm{1}−{c}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{u}}}= \\ $$$$=\left({k}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{4}} \mathrm{K}\:\left({c}^{\mathrm{2}} \right)\:\:\:\:\:\left[\mathrm{elliptic}\:\mathrm{K}\right] \\ $$$$\mathrm{K}\:\left({c}^{\mathrm{2}} \right)\:=\frac{\mathrm{2}}{\pi\mathrm{agm}\:\left(\mathrm{1},\:\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }\right)} \\ $$$${c}^{\mathrm{2}} =−\frac{\left(\mathrm{1}−\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\Omega=\frac{\mathrm{2}}{\pi\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} \mathrm{agm}\:\left(\mathrm{1},\:\frac{\mathrm{1}+\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} }{\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} }\right)} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{agm}\:\left({p},\:{q}\right)\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{v}_{{n}} \\ $$$${u}_{\mathrm{0}} ={p}\wedge{v}_{\mathrm{0}} ={q} \\ $$$${u}_{{n}+\mathrm{1}} =\frac{{u}_{{n}} +{v}_{{n}} }{\mathrm{2}}\wedge{v}_{{n}+\mathrm{1}} =\left({u}_{{n}} {v}_{{n}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$