Integrate $xoperatorname{erf}^{,3}(x),e^{-x^2},dx$












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 21:51







tgm

















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














  • 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










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