Integration of $-frac{32}{pi^2 k^2} int_k^{ksqrt2} a; L_2[a](sqrt{{a^2}/{k^2}-1}+arcsin(frac{k}{a});;) da$
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Hi I'm trying to solve this integral:
$I=-frac{32}{pi^2 k^2} int_k^{ksqrt2} a ;L_2[a](;;sqrt{frac{a^2}{k^2}-1}+arcsin(frac{k}{a});;) da$
where k is a parameter constrained by $0<k<1/sqrt2$ and $L_2[a]$ is the dilogarithm function.
Working on the integral, basically using by parts, I have reduced the integral in the form:
$I=-frac{1}{3}frac{80}{pi^2}L_2[ksqrt2]+frac{1}{3}frac{24}{pi}(L_2[k]-L_2[ksqrt2])\
;;;;;;-frac{1}{3}frac{48}{pi^2 k^2} int_k^{ksqrt2} a;ln(1-a);arcsin(frac{k}{a}); da\
;;;;;;-frac{1}{3}frac{16}{pi^2 k} int_k^{ksqrt2} ;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\
;;;;;;-frac{1}{3}frac{32}{pi^2 k^3} int_k^{ksqrt2} a^2;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\$
The three remaining integrals don't contain the dilogarithm function and they look simpler compared to the first one anyway I'm not able to find their closed forms. Another strategy I have used to solve the integral has been to notice that between brakets there is something similar to the integral of $asin(x)$, so I have rewritten the integral as:
$I=-frac{32}{pi^2 k^3} int_k^{ksqrt2} int_0^{k/a} a^2 L_2[a]; arcsin(x) dx; da\
;;;;;-frac{32}{pi^2 k^3}int_k^{ksqrt2}a^2 L_2[a];da$
In this case the second integral is solvable while the first double integral can be splitted in two parts dividing the domain of integration into a rectangular domain (and I have a solution for that) and in a triangle like domain (actually not a triangle ) and in this other case I find some other integrals similar to the integrals emerged with the first strategy.
Can someone give me some advice.
integration logarithms trigonometric-integrals polylogarithm
asked Nov 15 at 14:08
marco
383
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up vote
0
down vote
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Hi I'm trying to solve this integral:
$I=-frac{32}{pi^2 k^2} int_k^{ksqrt2} a ;L_2[a](;;sqrt{frac{a^2}{k^2}-1}+arcsin(frac{k}{a});;) da$
where k is a parameter constrained by $0<k<1/sqrt2$ and $L_2[a]$ is the dilogarithm function.
Working on the integral, basically using by parts, I have reduced the integral in the form:
$I=-frac{1}{3}frac{80}{pi^2}L_2[ksqrt2]+frac{1}{3}frac{24}{pi}(L_2[k]-L_2[ksqrt2])\
;;;;;;-frac{1}{3}frac{48}{pi^2 k^2} int_k^{ksqrt2} a;ln(1-a);arcsin(frac{k}{a}); da\
;;;;;;-frac{1}{3}frac{16}{pi^2 k} int_k^{ksqrt2} ;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\
;;;;;;-frac{1}{3}frac{32}{pi^2 k^3} int_k^{ksqrt2} a^2;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\$
The three remaining integrals don't contain the dilogarithm function and they look simpler compared to the first one anyway I'm not able to find their closed forms. Another strategy I have used to solve the integral has been to notice that between brakets there is something similar to the integral of $asin(x)$, so I have rewritten the integral as:
$I=-frac{32}{pi^2 k^3} int_k^{ksqrt2} int_0^{k/a} a^2 L_2[a]; arcsin(x) dx; da\
;;;;;-frac{32}{pi^2 k^3}int_k^{ksqrt2}a^2 L_2[a];da$
In this case the second integral is solvable while the first double integral can be splitted in two parts dividing the domain of integration into a rectangular domain (and I have a solution for that) and in a triangle like domain (actually not a triangle ) and in this other case I find some other integrals similar to the integrals emerged with the first strategy.
Can someone give me some advice.
integration logarithms trigonometric-integrals polylogarithm
asked Nov 15 at 14:08
marco
383
When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$,$arcsin(frac ka)$)?
– Henning Makholm
Nov 15 at 14:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hi I'm trying to solve this integral:
$I=-frac{32}{pi^2 k^2} int_k^{ksqrt2} a ;L_2[a](;;sqrt{frac{a^2}{k^2}-1}+arcsin(frac{k}{a});;) da$
where k is a parameter constrained by $0<k<1/sqrt2$ and $L_2[a]$ is the dilogarithm function.
Working on the integral, basically using by parts, I have reduced the integral in the form:
$I=-frac{1}{3}frac{80}{pi^2}L_2[ksqrt2]+frac{1}{3}frac{24}{pi}(L_2[k]-L_2[ksqrt2])\
;;;;;;-frac{1}{3}frac{48}{pi^2 k^2} int_k^{ksqrt2} a;ln(1-a);arcsin(frac{k}{a}); da\
;;;;;;-frac{1}{3}frac{16}{pi^2 k} int_k^{ksqrt2} ;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\
;;;;;;-frac{1}{3}frac{32}{pi^2 k^3} int_k^{ksqrt2} a^2;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\$
The three remaining integrals don't contain the dilogarithm function and they look simpler compared to the first one anyway I'm not able to find their closed forms. Another strategy I have used to solve the integral has been to notice that between brakets there is something similar to the integral of $asin(x)$, so I have rewritten the integral as:
$I=-frac{32}{pi^2 k^3} int_k^{ksqrt2} int_0^{k/a} a^2 L_2[a]; arcsin(x) dx; da\
;;;;;-frac{32}{pi^2 k^3}int_k^{ksqrt2}a^2 L_2[a];da$
In this case the second integral is solvable while the first double integral can be splitted in two parts dividing the domain of integration into a rectangular domain (and I have a solution for that) and in a triangle like domain (actually not a triangle ) and in this other case I find some other integrals similar to the integrals emerged with the first strategy.
Can someone give me some advice.
integration logarithms trigonometric-integrals polylogarithm
asked Nov 15 at 14:08
marco
383
Hi I'm trying to solve this integral:
$I=-frac{32}{pi^2 k^2} int_k^{ksqrt2} a ;L_2[a](;;sqrt{frac{a^2}{k^2}-1}+arcsin(frac{k}{a});;) da$
where k is a parameter constrained by $0<k<1/sqrt2$ and $L_2[a]$ is the dilogarithm function.
Working on the integral, basically using by parts, I have reduced the integral in the form:
$I=-frac{1}{3}frac{80}{pi^2}L_2[ksqrt2]+frac{1}{3}frac{24}{pi}(L_2[k]-L_2[ksqrt2])\
;;;;;;-frac{1}{3}frac{48}{pi^2 k^2} int_k^{ksqrt2} a;ln(1-a);arcsin(frac{k}{a}); da\
;;;;;;-frac{1}{3}frac{16}{pi^2 k} int_k^{ksqrt2} ;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\
;;;;;;-frac{1}{3}frac{32}{pi^2 k^3} int_k^{ksqrt2} a^2;ln(1-a);sqrt{1-frac{k^2}{a^2}}; da\$
The three remaining integrals don't contain the dilogarithm function and they look simpler compared to the first one anyway I'm not able to find their closed forms. Another strategy I have used to solve the integral has been to notice that between brakets there is something similar to the integral of $asin(x)$, so I have rewritten the integral as:
$I=-frac{32}{pi^2 k^3} int_k^{ksqrt2} int_0^{k/a} a^2 L_2[a]; arcsin(x) dx; da\
;;;;;-frac{32}{pi^2 k^3}int_k^{ksqrt2}a^2 L_2[a];da$
In this case the second integral is solvable while the first double integral can be splitted in two parts dividing the domain of integration into a rectangular domain (and I have a solution for that) and in a triangle like domain (actually not a triangle ) and in this other case I find some other integrals similar to the integrals emerged with the first strategy.
Can someone give me some advice.
integration logarithms trigonometric-integrals polylogarithm
integration logarithms trigonometric-integrals polylogarithm
asked Nov 15 at 14:08
marco
383
asked Nov 15 at 14:08
marco
383
edited Nov 15 at 15:55
asked Nov 15 at 14:08
marco
383
asked Nov 15 at 14:08
marco
383
asked Nov 15 at 14:08
marco
383
383
When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$,$arcsin(frac ka)$)?
– Henning Makholm
Nov 15 at 14:13
add a comment |
When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$,$arcsin(frac ka)$)?
– Henning Makholm
Nov 15 at 14:13
When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,
$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$, $arcsin(frac ka)$)?– Henning Makholm
Nov 15 at 14:13
When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,
$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$, $arcsin(frac ka)$)?– Henning Makholm
Nov 15 at 14:13
add a comment |
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When you write $asin(frac ka)$, do you mean $a$ times the sine of $frac ka$ (better written $asin(frac ka)$,
$asin(frac ka)$) or the arcsine of $frac ka$ (better written $arcsin(frac ka)$,$arcsin(frac ka)$)?– Henning Makholm
Nov 15 at 14:13