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How to solve the integral $int fracx^nsqrtx-x^2$?
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I am trying to find an appropriate substitution in the following indefinite integral
$int fracx^nsqrtx-x^2$
for an arbitrary power $n$ to obtainthe intermidiate steps to the same answer as in Mathematica expressed through the hypergeometric function as following
$$intfracx^ndxsqrtx(1-x)=-sqrt1-x;_2F_1left(frac12,frac12-n;frac32;1-xright)$$
Using this formula I could also compute the definite integral from $0$ to $1$, which is given as
$$int_0^1 fracx^ndxsqrtx(1-x)=fracsqrtpiGammaleft(n+frac12right)Gamma(n+1)$$
Any ideas are very welcome!
integration hypergeometric-function
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
asked Jul 16 at 21:46
Paul Frey
31
add a comment |Â
up vote
0
down vote
favorite
I am trying to find an appropriate substitution in the following indefinite integral
$int fracx^nsqrtx-x^2$
for an arbitrary power $n$ to obtainthe intermidiate steps to the same answer as in Mathematica expressed through the hypergeometric function as following
$$intfracx^ndxsqrtx(1-x)=-sqrt1-x;_2F_1left(frac12,frac12-n;frac32;1-xright)$$
Using this formula I could also compute the definite integral from $0$ to $1$, which is given as
$$int_0^1 fracx^ndxsqrtx(1-x)=fracsqrtpiGammaleft(n+frac12right)Gamma(n+1)$$
Any ideas are very welcome!
integration hypergeometric-function
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
asked Jul 16 at 21:46
Paul Frey
31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to find an appropriate substitution in the following indefinite integral
$int fracx^nsqrtx-x^2$
for an arbitrary power $n$ to obtainthe intermidiate steps to the same answer as in Mathematica expressed through the hypergeometric function as following
$$intfracx^ndxsqrtx(1-x)=-sqrt1-x;_2F_1left(frac12,frac12-n;frac32;1-xright)$$
Using this formula I could also compute the definite integral from $0$ to $1$, which is given as
$$int_0^1 fracx^ndxsqrtx(1-x)=fracsqrtpiGammaleft(n+frac12right)Gamma(n+1)$$
Any ideas are very welcome!
integration hypergeometric-function
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
asked Jul 16 at 21:46
Paul Frey
31
I am trying to find an appropriate substitution in the following indefinite integral
$int fracx^nsqrtx-x^2$
for an arbitrary power $n$ to obtainthe intermidiate steps to the same answer as in Mathematica expressed through the hypergeometric function as following
$$intfracx^ndxsqrtx(1-x)=-sqrt1-x;_2F_1left(frac12,frac12-n;frac32;1-xright)$$
Using this formula I could also compute the definite integral from $0$ to $1$, which is given as
$$int_0^1 fracx^ndxsqrtx(1-x)=fracsqrtpiGammaleft(n+frac12right)Gamma(n+1)$$
Any ideas are very welcome!
integration hypergeometric-function
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
asked Jul 16 at 21:46
Paul Frey
31
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
edited Jul 16 at 22:21
Thomas Andrews
128k10144285
128k10144285
asked Jul 16 at 21:46
Paul Frey
31
asked Jul 16 at 21:46
Paul Frey
31
asked Jul 16 at 21:46
Paul Frey
31
31
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If you only need the definite integral, then you can write it directly as the definition of the Beta function.
$$=int_0^1x^n-1/2(1-x)^-1/2=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=fracGamma(n+1/2)Gamma(1/2)Gamma(n+1/2+1/2)$$
Finally, use that $Gamma(1/2)=sqrtpi$.
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
add a comment |Â
up vote
0
down vote
Let $$x=sin^2(t)implies dx=2 sin (t) cos (t),dt$$ to make
$$int fracx^nsqrtx-x^2,dx=2int sin^2n(t),dt$$ and use the reduction formula (have a look here).
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If you only need the definite integral, then you can write it directly as the definition of the Beta function.
$$=int_0^1x^n-1/2(1-x)^-1/2=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=fracGamma(n+1/2)Gamma(1/2)Gamma(n+1/2+1/2)$$
Finally, use that $Gamma(1/2)=sqrtpi$.
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
add a comment |Â
up vote
3
down vote
accepted
If you only need the definite integral, then you can write it directly as the definition of the Beta function.
$$=int_0^1x^n-1/2(1-x)^-1/2=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=fracGamma(n+1/2)Gamma(1/2)Gamma(n+1/2+1/2)$$
Finally, use that $Gamma(1/2)=sqrtpi$.
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If you only need the definite integral, then you can write it directly as the definition of the Beta function.
$$=int_0^1x^n-1/2(1-x)^-1/2=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=fracGamma(n+1/2)Gamma(1/2)Gamma(n+1/2+1/2)$$
Finally, use that $Gamma(1/2)=sqrtpi$.
If you only need the definite integral, then you can write it directly as the definition of the Beta function.
$$=int_0^1x^n-1/2(1-x)^-1/2=B(n-1/2+1,-1/2+1)=B(n+1/2,1/2)=fracGamma(n+1/2)Gamma(1/2)Gamma(n+1/2+1/2)$$
Finally, use that $Gamma(1/2)=sqrtpi$.
answered Jul 16 at 22:07
user577471
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
add a comment |Â
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
Indeed, this can also be used to obtain a solution for the indefinite integral. One has to use the expression of the incomplete beta-function and the Chebychev integral $$int x^p(1-x)^qdx=B(x;1+p,1+q)$$ and then its relation to a hypergeometric function $$B(x;1+p,1+q)=fracx^1+p1+p _2F_1(1+p,-q;2+p;x)$$ Thanks a lot!
– Paul Frey
Jul 17 at 8:13
add a comment |Â
up vote
0
down vote
Let $$x=sin^2(t)implies dx=2 sin (t) cos (t),dt$$ to make
$$int fracx^nsqrtx-x^2,dx=2int sin^2n(t),dt$$ and use the reduction formula (have a look here).
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
add a comment |Â
up vote
0
down vote
Let $$x=sin^2(t)implies dx=2 sin (t) cos (t),dt$$ to make
$$int fracx^nsqrtx-x^2,dx=2int sin^2n(t),dt$$ and use the reduction formula (have a look here).
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $$x=sin^2(t)implies dx=2 sin (t) cos (t),dt$$ to make
$$int fracx^nsqrtx-x^2,dx=2int sin^2n(t),dt$$ and use the reduction formula (have a look here).
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
Let $$x=sin^2(t)implies dx=2 sin (t) cos (t),dt$$ to make
$$int fracx^nsqrtx-x^2,dx=2int sin^2n(t),dt$$ and use the reduction formula (have a look here).
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
answered Jul 17 at 3:36
Claude Leibovici
112k1055126
112k1055126
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
add a comment |Â
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
1
1
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
This approach does not help to proceed to the generalized formula which would be also expressed through a hypergeometric function. This brings me back to my original question...
– Paul Frey
Jul 17 at 7:10
add a comment |Â
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