Mixing
Integral of $sqrt{a(b(x+c)^2+1)}/((x-1)x^{3/2})$
3
$begingroup$
I am trying to solve
$$int_{x_0}^inftyfrac{sqrt{a(b(x+c)^2+1)}}{(x-1)x^{3/2}}dx$$
where, $a,b,c,x_0,xinBbb R$ and $b,c,x>0$ and $x_0>1$.
I tried a $sinh$ substitution without too much success. I also tried something like $v=sqrt{a(b(x+c)^2+1)}quad$, but maybe I should first do some more transformations before attempting something like that.
This is not homework. I would be grateful for some hints.
integration definite-integrals radicals rational-functions
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
$endgroup$
|
show 4 more comments
3
$begingroup$
I am trying to solve
$$int_{x_0}^inftyfrac{sqrt{a(b(x+c)^2+1)}}{(x-1)x^{3/2}}dx$$
where, $a,b,c,x_0,xinBbb R$ and $b,c,x>0$ and $x_0>1$.
I tried a $sinh$ substitution without too much success. I also tried something like $v=sqrt{a(b(x+c)^2+1)}quad$, but maybe I should first do some more transformations before attempting something like that.
This is not homework. I would be grateful for some hints.
integration definite-integrals radicals rational-functions
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
$endgroup$
$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
1
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
1
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33
|
show 4 more comments
3
3
3
1
$begingroup$
I am trying to solve
$$int_{x_0}^inftyfrac{sqrt{a(b(x+c)^2+1)}}{(x-1)x^{3/2}}dx$$
where, $a,b,c,x_0,xinBbb R$ and $b,c,x>0$ and $x_0>1$.
I tried a $sinh$ substitution without too much success. I also tried something like $v=sqrt{a(b(x+c)^2+1)}quad$, but maybe I should first do some more transformations before attempting something like that.
This is not homework. I would be grateful for some hints.
integration definite-integrals radicals rational-functions
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
$endgroup$
I am trying to solve
$$int_{x_0}^inftyfrac{sqrt{a(b(x+c)^2+1)}}{(x-1)x^{3/2}}dx$$
where, $a,b,c,x_0,xinBbb R$ and $b,c,x>0$ and $x_0>1$.
I tried a $sinh$ substitution without too much success. I also tried something like $v=sqrt{a(b(x+c)^2+1)}quad$, but maybe I should first do some more transformations before attempting something like that.
This is not homework. I would be grateful for some hints.
integration definite-integrals radicals rational-functions
integration definite-integrals radicals rational-functions
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
edited Jan 31 at 14:25
ThunderBiggi
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
asked Jan 31 at 12:51
ThunderBiggiThunderBiggi
1508
1508
$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
1
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
1
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33
|
show 4 more comments
$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
1
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
1
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33
$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
1
1
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
1
1
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33
|
show 4 more comments
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$begingroup$
If $x_0 leq 1$ the integral is divergent.
$endgroup$
– PierreCarre
Jan 31 at 13:38
1
$begingroup$
The primitive comes in terms of elliptic functions... I don't think you'll be able to get an expression with only "ususl" functions.
$endgroup$
– PierreCarre
Jan 31 at 13:45
$begingroup$
@Zacky I had a quick look at that case and it doesn't appear to be much simpler.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
$begingroup$
@PierreCarre Yes, I forgot to include that $x_0>1$. I got an answer full of Elliptic functions from Mathematica, but that is not useful to me. I guess I will have to try and play with it.
$endgroup$
– ThunderBiggi
Jan 31 at 14:25
1
$begingroup$
@ThunderBiggi, don't bother playing with it... If you were able to get an answer not involving elliptic functions then you would have proved that elliptic functions can be written in closed form in terms of "usual" functions, which is not the case.
$endgroup$
– PierreCarre
Jan 31 at 14:33