Improper integral containing $sqrt{cos x-frac1{sqrt 2}}$ in the denominator











up vote
8
down vote

favorite
6












How do I find the value of this integral--



$$I=displaystyleint_{0}^{pi/4} frac{sec^2 x dx}{sqrt {cos x-dfrac{1}{sqrt 2}}}$$



I came across this integral too in physics.






integration definite-integrals improper-integrals







share|cite|improve this question
























  • The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
    – Claude Leibovici
    Aug 20 '14 at 18:39












  • @ClaudeLeibovici How do we do it then?
    – pkwssis
    Aug 20 '14 at 18:40












  • I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
    – Claude Leibovici
    Aug 20 '14 at 18:43










  • @ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
    – pkwssis
    Aug 20 '14 at 18:45












  • The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
    – user111187
    Aug 20 '14 at 19:00















up vote
8
down vote

favorite
6












How do I find the value of this integral--



$$I=displaystyleint_{0}^{pi/4} frac{sec^2 x dx}{sqrt {cos x-dfrac{1}{sqrt 2}}}$$



I came across this integral too in physics.






integration definite-integrals improper-integrals







share|cite|improve this question
























  • The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
    – Claude Leibovici
    Aug 20 '14 at 18:39












  • @ClaudeLeibovici How do we do it then?
    – pkwssis
    Aug 20 '14 at 18:40












  • I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
    – Claude Leibovici
    Aug 20 '14 at 18:43










  • @ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
    – pkwssis
    Aug 20 '14 at 18:45












  • The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
    – user111187
    Aug 20 '14 at 19:00













up vote
8
down vote

favorite
6









up vote
8
down vote

favorite
6






6





How do I find the value of this integral--



$$I=displaystyleint_{0}^{pi/4} frac{sec^2 x dx}{sqrt {cos x-dfrac{1}{sqrt 2}}}$$



I came across this integral too in physics.






integration definite-integrals improper-integrals







share|cite|improve this question















How do I find the value of this integral--



$$I=displaystyleint_{0}^{pi/4} frac{sec^2 x dx}{sqrt {cos x-dfrac{1}{sqrt 2}}}$$



I came across this integral too in physics.






integration definite-integrals improper-integrals




integration definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 12:18









Martin Sleziak

44.6k7115269




44.6k7115269










asked Aug 20 '14 at 18:18









pkwssis

2,0441333




2,0441333












  • The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
    – Claude Leibovici
    Aug 20 '14 at 18:39












  • @ClaudeLeibovici How do we do it then?
    – pkwssis
    Aug 20 '14 at 18:40












  • I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
    – Claude Leibovici
    Aug 20 '14 at 18:43










  • @ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
    – pkwssis
    Aug 20 '14 at 18:45












  • The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
    – user111187
    Aug 20 '14 at 19:00


















  • The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
    – Claude Leibovici
    Aug 20 '14 at 18:39












  • @ClaudeLeibovici How do we do it then?
    – pkwssis
    Aug 20 '14 at 18:40












  • I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
    – Claude Leibovici
    Aug 20 '14 at 18:43










  • @ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
    – pkwssis
    Aug 20 '14 at 18:45












  • The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
    – user111187
    Aug 20 '14 at 19:00
















The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
– Claude Leibovici
Aug 20 '14 at 18:39






The antiderivative does exist ! The fact that it is a monster is another story ! Cheers.
– Claude Leibovici
Aug 20 '14 at 18:39














@ClaudeLeibovici How do we do it then?
– pkwssis
Aug 20 '14 at 18:40






@ClaudeLeibovici How do we do it then?
– pkwssis
Aug 20 '14 at 18:40














I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
– Claude Leibovici
Aug 20 '14 at 18:43




I did not do it ! A CAS did it but it is really a monster; the same for the integral ($3.338953436$)
– Claude Leibovici
Aug 20 '14 at 18:43












@ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
– pkwssis
Aug 20 '14 at 18:45






@ClaudeLeibovici Do you know the closed form? I don't have mathematica or similar software and wolfram alpha doesn't provide the closed form
– pkwssis
Aug 20 '14 at 18:45














The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
– user111187
Aug 20 '14 at 19:00




The closed form is apparently $$frac{8 Gamma left(-frac{1}{4}right) Gamma left(frac{7}{4}right) , _3F_2left(frac{3}{4},frac{5}{4},frac{7}{4};frac{3}{2},frac{9}{4};frac{1}{2}right)-5 sqrt{2} Gamma left(-frac{3}{4}right) Gamma left(frac{1}{4}right) , _3F_2left(frac{1}{4},frac{3}{4},frac{5}{4};frac{1}{2},frac{7}{4};frac{1}{2}right)}{80 sqrt{pi }}+frac{pi }{sqrt[4]{2}}$$, but the approximate numerical value is much more useful.
– user111187
Aug 20 '14 at 19:00










2 Answers
2






active

oldest

votes

















up vote
12
down vote



accepted










$$I=fracpi{sqrt[4]2}+sqrt{20-14sqrt2} K!left(2sqrt2-3right)+sqrt{2+sqrt2} E!left(2sqrt2-3right)\-2sqrt{2-sqrt2} Pileft(sqrt2-1,,2sqrt2-3right)$$
where $K(m), E(m), Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind:
$$K(m)={largeint}_0^{pi/2}frac{dtheta}{sqrt{1-msin^2theta}}$$
$$E(m)={largeint}_0^{pi/2}sqrt{1-msin^2theta},dtheta$$
$$Pi(n,m)={Largeint}_0^{pi/2}frac{dtheta}{left(1-nsin^2thetaright)sqrt {1-msin^2theta}}$$
Note that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.






share|cite|improve this answer

















  • 1




    It would be greatly appreciated if you could show the steps. Thanks! :)
    – Pranav Arora
    Aug 22 '14 at 18:59




















up vote
6
down vote













With $t=cos(x)$, we see it is an elliptic integral,
$$
I = int_{1/sqrt{2}}^1 frac{dt}{t^2sqrt{(1-t^2)(t-1/sqrt{2})}}
$$



added



Maple gets: if $0<a<1$, then
$$
int _{a}^{1}!{frac {dt}{{t}^{2}sqrt { left( 1-t^2 right)
left( t-a right) }}}{}
=-{frac {sqrt {2}{bf K} left( 1
/2,sqrt {-2,a+2} right) }{a}}
+{frac {sqrt {2} left( a+1
right) {bf E} left( 1/2,sqrt {-2,a+2} right) }{a
left( 1+a right) }}
+frac{1}{sqrt {2}} left( a+1 right) {bf Pi} left({frac {a-1}{2a}},
frac{sqrt {-2,a+2}}{2} right)
{a}^{-2}
$$
and in particular with $a=1/sqrt{2}$,
$$
I = 2,{bf E} left( 1/2,sqrt {2-sqrt {2}} right) -2,{bf K} left( 1/2,sqrt {2-sqrt {2}} right) + left( 1+sqrt {2
} right) {bf Pi} left( 1/4, left( -2+sqrt {2} right)
sqrt {2},1/2,sqrt {2-sqrt {2}} right) approx 3.338954
$$
where Maple's notation uses the modulus $k$.






share|cite|improve this answer



















  • 1




    I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
    – Liu Jin Tsai
    Aug 21 '14 at 21:12











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
12
down vote



accepted










$$I=fracpi{sqrt[4]2}+sqrt{20-14sqrt2} K!left(2sqrt2-3right)+sqrt{2+sqrt2} E!left(2sqrt2-3right)\-2sqrt{2-sqrt2} Pileft(sqrt2-1,,2sqrt2-3right)$$
where $K(m), E(m), Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind:
$$K(m)={largeint}_0^{pi/2}frac{dtheta}{sqrt{1-msin^2theta}}$$
$$E(m)={largeint}_0^{pi/2}sqrt{1-msin^2theta},dtheta$$
$$Pi(n,m)={Largeint}_0^{pi/2}frac{dtheta}{left(1-nsin^2thetaright)sqrt {1-msin^2theta}}$$
Note that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.






share|cite|improve this answer

















  • 1




    It would be greatly appreciated if you could show the steps. Thanks! :)
    – Pranav Arora
    Aug 22 '14 at 18:59

















up vote
12
down vote



accepted










$$I=fracpi{sqrt[4]2}+sqrt{20-14sqrt2} K!left(2sqrt2-3right)+sqrt{2+sqrt2} E!left(2sqrt2-3right)\-2sqrt{2-sqrt2} Pileft(sqrt2-1,,2sqrt2-3right)$$
where $K(m), E(m), Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind:
$$K(m)={largeint}_0^{pi/2}frac{dtheta}{sqrt{1-msin^2theta}}$$
$$E(m)={largeint}_0^{pi/2}sqrt{1-msin^2theta},dtheta$$
$$Pi(n,m)={Largeint}_0^{pi/2}frac{dtheta}{left(1-nsin^2thetaright)sqrt {1-msin^2theta}}$$
Note that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.






share|cite|improve this answer

















  • 1




    It would be greatly appreciated if you could show the steps. Thanks! :)
    – Pranav Arora
    Aug 22 '14 at 18:59















up vote
12
down vote



accepted







up vote
12
down vote



accepted






$$I=fracpi{sqrt[4]2}+sqrt{20-14sqrt2} K!left(2sqrt2-3right)+sqrt{2+sqrt2} E!left(2sqrt2-3right)\-2sqrt{2-sqrt2} Pileft(sqrt2-1,,2sqrt2-3right)$$
where $K(m), E(m), Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind:
$$K(m)={largeint}_0^{pi/2}frac{dtheta}{sqrt{1-msin^2theta}}$$
$$E(m)={largeint}_0^{pi/2}sqrt{1-msin^2theta},dtheta$$
$$Pi(n,m)={Largeint}_0^{pi/2}frac{dtheta}{left(1-nsin^2thetaright)sqrt {1-msin^2theta}}$$
Note that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.






share|cite|improve this answer












$$I=fracpi{sqrt[4]2}+sqrt{20-14sqrt2} K!left(2sqrt2-3right)+sqrt{2+sqrt2} E!left(2sqrt2-3right)\-2sqrt{2-sqrt2} Pileft(sqrt2-1,,2sqrt2-3right)$$
where $K(m), E(m), Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind:
$$K(m)={largeint}_0^{pi/2}frac{dtheta}{sqrt{1-msin^2theta}}$$
$$E(m)={largeint}_0^{pi/2}sqrt{1-msin^2theta},dtheta$$
$$Pi(n,m)={Largeint}_0^{pi/2}frac{dtheta}{left(1-nsin^2thetaright)sqrt {1-msin^2theta}}$$
Note that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 21 '14 at 2:19









Cleo

12.8k43560




12.8k43560








  • 1




    It would be greatly appreciated if you could show the steps. Thanks! :)
    – Pranav Arora
    Aug 22 '14 at 18:59
















  • 1




    It would be greatly appreciated if you could show the steps. Thanks! :)
    – Pranav Arora
    Aug 22 '14 at 18:59










1




1




It would be greatly appreciated if you could show the steps. Thanks! :)
– Pranav Arora
Aug 22 '14 at 18:59






It would be greatly appreciated if you could show the steps. Thanks! :)
– Pranav Arora
Aug 22 '14 at 18:59












up vote
6
down vote













With $t=cos(x)$, we see it is an elliptic integral,
$$
I = int_{1/sqrt{2}}^1 frac{dt}{t^2sqrt{(1-t^2)(t-1/sqrt{2})}}
$$



added



Maple gets: if $0<a<1$, then
$$
int _{a}^{1}!{frac {dt}{{t}^{2}sqrt { left( 1-t^2 right)
left( t-a right) }}}{}
=-{frac {sqrt {2}{bf K} left( 1
/2,sqrt {-2,a+2} right) }{a}}
+{frac {sqrt {2} left( a+1
right) {bf E} left( 1/2,sqrt {-2,a+2} right) }{a
left( 1+a right) }}
+frac{1}{sqrt {2}} left( a+1 right) {bf Pi} left({frac {a-1}{2a}},
frac{sqrt {-2,a+2}}{2} right)
{a}^{-2}
$$
and in particular with $a=1/sqrt{2}$,
$$
I = 2,{bf E} left( 1/2,sqrt {2-sqrt {2}} right) -2,{bf K} left( 1/2,sqrt {2-sqrt {2}} right) + left( 1+sqrt {2
} right) {bf Pi} left( 1/4, left( -2+sqrt {2} right)
sqrt {2},1/2,sqrt {2-sqrt {2}} right) approx 3.338954
$$
where Maple's notation uses the modulus $k$.






share|cite|improve this answer



















  • 1




    I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
    – Liu Jin Tsai
    Aug 21 '14 at 21:12















up vote
6
down vote













With $t=cos(x)$, we see it is an elliptic integral,
$$
I = int_{1/sqrt{2}}^1 frac{dt}{t^2sqrt{(1-t^2)(t-1/sqrt{2})}}
$$



added



Maple gets: if $0<a<1$, then
$$
int _{a}^{1}!{frac {dt}{{t}^{2}sqrt { left( 1-t^2 right)
left( t-a right) }}}{}
=-{frac {sqrt {2}{bf K} left( 1
/2,sqrt {-2,a+2} right) }{a}}
+{frac {sqrt {2} left( a+1
right) {bf E} left( 1/2,sqrt {-2,a+2} right) }{a
left( 1+a right) }}
+frac{1}{sqrt {2}} left( a+1 right) {bf Pi} left({frac {a-1}{2a}},
frac{sqrt {-2,a+2}}{2} right)
{a}^{-2}
$$
and in particular with $a=1/sqrt{2}$,
$$
I = 2,{bf E} left( 1/2,sqrt {2-sqrt {2}} right) -2,{bf K} left( 1/2,sqrt {2-sqrt {2}} right) + left( 1+sqrt {2
} right) {bf Pi} left( 1/4, left( -2+sqrt {2} right)
sqrt {2},1/2,sqrt {2-sqrt {2}} right) approx 3.338954
$$
where Maple's notation uses the modulus $k$.






share|cite|improve this answer



















  • 1




    I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
    – Liu Jin Tsai
    Aug 21 '14 at 21:12













up vote
6
down vote










up vote
6
down vote









With $t=cos(x)$, we see it is an elliptic integral,
$$
I = int_{1/sqrt{2}}^1 frac{dt}{t^2sqrt{(1-t^2)(t-1/sqrt{2})}}
$$



added



Maple gets: if $0<a<1$, then
$$
int _{a}^{1}!{frac {dt}{{t}^{2}sqrt { left( 1-t^2 right)
left( t-a right) }}}{}
=-{frac {sqrt {2}{bf K} left( 1
/2,sqrt {-2,a+2} right) }{a}}
+{frac {sqrt {2} left( a+1
right) {bf E} left( 1/2,sqrt {-2,a+2} right) }{a
left( 1+a right) }}
+frac{1}{sqrt {2}} left( a+1 right) {bf Pi} left({frac {a-1}{2a}},
frac{sqrt {-2,a+2}}{2} right)
{a}^{-2}
$$
and in particular with $a=1/sqrt{2}$,
$$
I = 2,{bf E} left( 1/2,sqrt {2-sqrt {2}} right) -2,{bf K} left( 1/2,sqrt {2-sqrt {2}} right) + left( 1+sqrt {2
} right) {bf Pi} left( 1/4, left( -2+sqrt {2} right)
sqrt {2},1/2,sqrt {2-sqrt {2}} right) approx 3.338954
$$
where Maple's notation uses the modulus $k$.






share|cite|improve this answer














With $t=cos(x)$, we see it is an elliptic integral,
$$
I = int_{1/sqrt{2}}^1 frac{dt}{t^2sqrt{(1-t^2)(t-1/sqrt{2})}}
$$



added



Maple gets: if $0<a<1$, then
$$
int _{a}^{1}!{frac {dt}{{t}^{2}sqrt { left( 1-t^2 right)
left( t-a right) }}}{}
=-{frac {sqrt {2}{bf K} left( 1
/2,sqrt {-2,a+2} right) }{a}}
+{frac {sqrt {2} left( a+1
right) {bf E} left( 1/2,sqrt {-2,a+2} right) }{a
left( 1+a right) }}
+frac{1}{sqrt {2}} left( a+1 right) {bf Pi} left({frac {a-1}{2a}},
frac{sqrt {-2,a+2}}{2} right)
{a}^{-2}
$$
and in particular with $a=1/sqrt{2}$,
$$
I = 2,{bf E} left( 1/2,sqrt {2-sqrt {2}} right) -2,{bf K} left( 1/2,sqrt {2-sqrt {2}} right) + left( 1+sqrt {2
} right) {bf Pi} left( 1/4, left( -2+sqrt {2} right)
sqrt {2},1/2,sqrt {2-sqrt {2}} right) approx 3.338954
$$
where Maple's notation uses the modulus $k$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 21 '14 at 22:18

























answered Aug 20 '14 at 19:10









GEdgar

61.3k267167




61.3k267167








  • 1




    I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
    – Liu Jin Tsai
    Aug 21 '14 at 21:12














  • 1




    I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
    – Liu Jin Tsai
    Aug 21 '14 at 21:12








1




1




I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
– Liu Jin Tsai
Aug 21 '14 at 21:12




I do not expect Cleo to post a proof. Have you got the same closed form as hers? Can you give more details?
– Liu Jin Tsai
Aug 21 '14 at 21:12


















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