Mixing
Integrate $xoperatorname{erf}^{,3}(x),e^{-x^2},dx$
$begingroup$
Looking for a way to perform this integral related to the error function.
I am thinking an answer in closed form cannot be done, but hoping I missed something.
$$
int xoperatorname{erf}^{,3}(x),e^{-x^2}dx
$$
Edit:
Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have
$int u,dv = uv-int v,du$
$u=x,operatorname{erf}(x)$
$du=operatorname{erf}(x)+frac{2x}{sqrt{pi}}e^{-x^2}dx$
$v=frac{pi}{6}operatorname{erf}^3(x)$
$dv=operatorname{erf}^2(x),e^{-x^2}dx$
$int xoperatorname{erf}^{,3}(x),e^{-x^2},dx = frac{sqrt{pi}}{6},x,e^{-x^2}operatorname{erf}^{,3}(x)-frac{sqrt{pi}}{6},int operatorname{erf}^3(x)[operatorname{erf}(x)+frac{2}{sqrt{pi}}xe^{-x^2}]dx$
$=frac{sqrt{pi}}{4}xoperatorname{erf}^4(x)-frac{sqrt{pi}}{4}int operatorname{erf}^{,4}(x)dx$
Then the challenge becomes how to solve:
$$
int operatorname{erf}^{,4}(x),dx
$$
Edit 2:
Able to simplify(?) a bit more:
$int operatorname{erf}^{,4}(x),dx = xoperatorname{erf}^{,4}(x)+frac{4}{sqrt{pi}}e^{-x^2}operatorname{erf}^{,3}(x)-frac{24}{pi}int operatorname{erf}^{,2}(x)e^{-2x^2}dx$
Now the challenge becomes how to solve:
$$
int operatorname{erf}^{,2}(x)e^{-2x^2}dx
$$
calculus integration indefinite-integrals error-function
asked Jan 10 at 7:39
tgmtgm
61
$endgroup$
add a comment |
$begingroup$
Looking for a way to perform this integral related to the error function.
I am thinking an answer in closed form cannot be done, but hoping I missed something.
$$
int xoperatorname{erf}^{,3}(x),e^{-x^2}dx
$$
Edit:
Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have
$int u,dv = uv-int v,du$
$u=x,operatorname{erf}(x)$
$du=operatorname{erf}(x)+frac{2x}{sqrt{pi}}e^{-x^2}dx$
$v=frac{pi}{6}operatorname{erf}^3(x)$
$dv=operatorname{erf}^2(x),e^{-x^2}dx$
$int xoperatorname{erf}^{,3}(x),e^{-x^2},dx = frac{sqrt{pi}}{6},x,e^{-x^2}operatorname{erf}^{,3}(x)-frac{sqrt{pi}}{6},int operatorname{erf}^3(x)[operatorname{erf}(x)+frac{2}{sqrt{pi}}xe^{-x^2}]dx$
$=frac{sqrt{pi}}{4}xoperatorname{erf}^4(x)-frac{sqrt{pi}}{4}int operatorname{erf}^{,4}(x)dx$
Then the challenge becomes how to solve:
$$
int operatorname{erf}^{,4}(x),dx
$$
Edit 2:
Able to simplify(?) a bit more:
$int operatorname{erf}^{,4}(x),dx = xoperatorname{erf}^{,4}(x)+frac{4}{sqrt{pi}}e^{-x^2}operatorname{erf}^{,3}(x)-frac{24}{pi}int operatorname{erf}^{,2}(x)e^{-2x^2}dx$
Now the challenge becomes how to solve:
$$
int operatorname{erf}^{,2}(x)e^{-2x^2}dx
$$
calculus integration indefinite-integrals error-function
asked Jan 10 at 7:39
tgmtgm
61
$endgroup$
1
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13
add a comment |
$begingroup$
Looking for a way to perform this integral related to the error function.
I am thinking an answer in closed form cannot be done, but hoping I missed something.
$$
int xoperatorname{erf}^{,3}(x),e^{-x^2}dx
$$
Edit:
Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have
$int u,dv = uv-int v,du$
$u=x,operatorname{erf}(x)$
$du=operatorname{erf}(x)+frac{2x}{sqrt{pi}}e^{-x^2}dx$
$v=frac{pi}{6}operatorname{erf}^3(x)$
$dv=operatorname{erf}^2(x),e^{-x^2}dx$
$int xoperatorname{erf}^{,3}(x),e^{-x^2},dx = frac{sqrt{pi}}{6},x,e^{-x^2}operatorname{erf}^{,3}(x)-frac{sqrt{pi}}{6},int operatorname{erf}^3(x)[operatorname{erf}(x)+frac{2}{sqrt{pi}}xe^{-x^2}]dx$
$=frac{sqrt{pi}}{4}xoperatorname{erf}^4(x)-frac{sqrt{pi}}{4}int operatorname{erf}^{,4}(x)dx$
Then the challenge becomes how to solve:
$$
int operatorname{erf}^{,4}(x),dx
$$
Edit 2:
Able to simplify(?) a bit more:
$int operatorname{erf}^{,4}(x),dx = xoperatorname{erf}^{,4}(x)+frac{4}{sqrt{pi}}e^{-x^2}operatorname{erf}^{,3}(x)-frac{24}{pi}int operatorname{erf}^{,2}(x)e^{-2x^2}dx$
Now the challenge becomes how to solve:
$$
int operatorname{erf}^{,2}(x)e^{-2x^2}dx
$$
calculus integration indefinite-integrals error-function
asked Jan 10 at 7:39
tgmtgm
61
$endgroup$
Looking for a way to perform this integral related to the error function.
I am thinking an answer in closed form cannot be done, but hoping I missed something.
$$
int xoperatorname{erf}^{,3}(x),e^{-x^2}dx
$$
Edit:
Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have
$int u,dv = uv-int v,du$
$u=x,operatorname{erf}(x)$
$du=operatorname{erf}(x)+frac{2x}{sqrt{pi}}e^{-x^2}dx$
$v=frac{pi}{6}operatorname{erf}^3(x)$
$dv=operatorname{erf}^2(x),e^{-x^2}dx$
$int xoperatorname{erf}^{,3}(x),e^{-x^2},dx = frac{sqrt{pi}}{6},x,e^{-x^2}operatorname{erf}^{,3}(x)-frac{sqrt{pi}}{6},int operatorname{erf}^3(x)[operatorname{erf}(x)+frac{2}{sqrt{pi}}xe^{-x^2}]dx$
$=frac{sqrt{pi}}{4}xoperatorname{erf}^4(x)-frac{sqrt{pi}}{4}int operatorname{erf}^{,4}(x)dx$
Then the challenge becomes how to solve:
$$
int operatorname{erf}^{,4}(x),dx
$$
Edit 2:
Able to simplify(?) a bit more:
$int operatorname{erf}^{,4}(x),dx = xoperatorname{erf}^{,4}(x)+frac{4}{sqrt{pi}}e^{-x^2}operatorname{erf}^{,3}(x)-frac{24}{pi}int operatorname{erf}^{,2}(x)e^{-2x^2}dx$
Now the challenge becomes how to solve:
$$
int operatorname{erf}^{,2}(x)e^{-2x^2}dx
$$
calculus integration indefinite-integrals error-function
calculus integration indefinite-integrals error-function
asked Jan 10 at 7:39
tgmtgm
61
asked Jan 10 at 7:39
tgmtgm
61
edited Jan 10 at 21:51
tgm
asked Jan 10 at 7:39
tgmtgm
61
asked Jan 10 at 7:39
tgmtgm
61
asked Jan 10 at 7:39
tgmtgm
61
61
1
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13
add a comment |
1
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13
1
1
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13
add a comment |
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1
$begingroup$
Integration by parts can reduce the integrand to $text{erf }^4(x)$, indeed not manageable.
$endgroup$
– Yves Daoust
Jan 10 at 11:22
$begingroup$
Where did you find this monster?
$endgroup$
– Yuriy S
Jan 10 at 11:30
$begingroup$
Let $u=erf(x)$ and all is good.
$endgroup$
– B. Goddard
Jan 10 at 12:06
$begingroup$
@B.Goddard: can you elaborate ?
$endgroup$
– Yves Daoust
Jan 10 at 12:40
$begingroup$
If that $x$ wasn't there it'd be a lot easier.
$endgroup$
– Michael Seifert
Jan 10 at 21:13