An $operatorname{erfi}(x)e^{-x^2}$ integral












8












$begingroup$


I want to find an elementary evaluation of




$$I=int_0^infty left(frac{sqrtpi}2operatorname{erfi}(x)e^{-x^2}-frac1{1+2x}right)dx$$
where $operatorname{erfi}(x)=frac{2}{sqrtpi}int_0^xe^{t^2}dt$.




Rough Solution

$$I=int_0^inftyleft({}_1F_1(1;3/2,-x^2)x-frac1{1+2x}right)dx$$
$$=left(frac{x^2}2{}_2F_2(1,1;3/2,2,-x^2)-frac12ln(1+2x)right)Bigg|_0^infty$$
By using the asymptotic expansion of $_2F_2$ I can get the answer is $frac{gamma}4$, where $gamma$ is the Euler's constant.

I wonder if there is a elementary proof without using hypergeometric function.






calculus integration definite-integrals special-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
    $endgroup$
    – Riemann
    Sep 14 '18 at 9:25












  • $begingroup$
    @Riemann. According to the result, it is $text{erfi}$
    $endgroup$
    – Claude Leibovici
    Sep 14 '18 at 9:57






  • 1




    $begingroup$
    Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
    $endgroup$
    – Batominovski
    Sep 14 '18 at 10:19












  • $begingroup$
    @KemonoChen Can you please clarify your question.
    $endgroup$
    – paulplusx
    Sep 14 '18 at 13:16










  • $begingroup$
    It is $operatorname{erfi}$. Sorry for the typo.
    $endgroup$
    – Kemono Chen
    Sep 14 '18 at 13:16
















8












$begingroup$


I want to find an elementary evaluation of




$$I=int_0^infty left(frac{sqrtpi}2operatorname{erfi}(x)e^{-x^2}-frac1{1+2x}right)dx$$
where $operatorname{erfi}(x)=frac{2}{sqrtpi}int_0^xe^{t^2}dt$.




Rough Solution

$$I=int_0^inftyleft({}_1F_1(1;3/2,-x^2)x-frac1{1+2x}right)dx$$
$$=left(frac{x^2}2{}_2F_2(1,1;3/2,2,-x^2)-frac12ln(1+2x)right)Bigg|_0^infty$$
By using the asymptotic expansion of $_2F_2$ I can get the answer is $frac{gamma}4$, where $gamma$ is the Euler's constant.

I wonder if there is a elementary proof without using hypergeometric function.






calculus integration definite-integrals special-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
    $endgroup$
    – Riemann
    Sep 14 '18 at 9:25












  • $begingroup$
    @Riemann. According to the result, it is $text{erfi}$
    $endgroup$
    – Claude Leibovici
    Sep 14 '18 at 9:57






  • 1




    $begingroup$
    Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
    $endgroup$
    – Batominovski
    Sep 14 '18 at 10:19












  • $begingroup$
    @KemonoChen Can you please clarify your question.
    $endgroup$
    – paulplusx
    Sep 14 '18 at 13:16










  • $begingroup$
    It is $operatorname{erfi}$. Sorry for the typo.
    $endgroup$
    – Kemono Chen
    Sep 14 '18 at 13:16














8












8








8


4



$begingroup$


I want to find an elementary evaluation of




$$I=int_0^infty left(frac{sqrtpi}2operatorname{erfi}(x)e^{-x^2}-frac1{1+2x}right)dx$$
where $operatorname{erfi}(x)=frac{2}{sqrtpi}int_0^xe^{t^2}dt$.




Rough Solution

$$I=int_0^inftyleft({}_1F_1(1;3/2,-x^2)x-frac1{1+2x}right)dx$$
$$=left(frac{x^2}2{}_2F_2(1,1;3/2,2,-x^2)-frac12ln(1+2x)right)Bigg|_0^infty$$
By using the asymptotic expansion of $_2F_2$ I can get the answer is $frac{gamma}4$, where $gamma$ is the Euler's constant.

I wonder if there is a elementary proof without using hypergeometric function.






calculus integration definite-integrals special-functions







share|cite|improve this question











$endgroup$




I want to find an elementary evaluation of




$$I=int_0^infty left(frac{sqrtpi}2operatorname{erfi}(x)e^{-x^2}-frac1{1+2x}right)dx$$
where $operatorname{erfi}(x)=frac{2}{sqrtpi}int_0^xe^{t^2}dt$.




Rough Solution

$$I=int_0^inftyleft({}_1F_1(1;3/2,-x^2)x-frac1{1+2x}right)dx$$
$$=left(frac{x^2}2{}_2F_2(1,1;3/2,2,-x^2)-frac12ln(1+2x)right)Bigg|_0^infty$$
By using the asymptotic expansion of $_2F_2$ I can get the answer is $frac{gamma}4$, where $gamma$ is the Euler's constant.

I wonder if there is a elementary proof without using hypergeometric function.






calculus integration definite-integrals special-functions




calculus integration definite-integrals special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 14 '18 at 13:15







Kemono Chen

















asked Sep 14 '18 at 8:43









Kemono ChenKemono Chen

3,2021844




3,2021844








  • 2




    $begingroup$
    $erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
    $endgroup$
    – Riemann
    Sep 14 '18 at 9:25












  • $begingroup$
    @Riemann. According to the result, it is $text{erfi}$
    $endgroup$
    – Claude Leibovici
    Sep 14 '18 at 9:57






  • 1




    $begingroup$
    Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
    $endgroup$
    – Batominovski
    Sep 14 '18 at 10:19












  • $begingroup$
    @KemonoChen Can you please clarify your question.
    $endgroup$
    – paulplusx
    Sep 14 '18 at 13:16










  • $begingroup$
    It is $operatorname{erfi}$. Sorry for the typo.
    $endgroup$
    – Kemono Chen
    Sep 14 '18 at 13:16














  • 2




    $begingroup$
    $erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
    $endgroup$
    – Riemann
    Sep 14 '18 at 9:25












  • $begingroup$
    @Riemann. According to the result, it is $text{erfi}$
    $endgroup$
    – Claude Leibovici
    Sep 14 '18 at 9:57






  • 1




    $begingroup$
    Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
    $endgroup$
    – Batominovski
    Sep 14 '18 at 10:19












  • $begingroup$
    @KemonoChen Can you please clarify your question.
    $endgroup$
    – paulplusx
    Sep 14 '18 at 13:16










  • $begingroup$
    It is $operatorname{erfi}$. Sorry for the typo.
    $endgroup$
    – Kemono Chen
    Sep 14 '18 at 13:16








2




2




$begingroup$
$erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
$endgroup$
– Riemann
Sep 14 '18 at 9:25






$begingroup$
$erfi(x)$ and $erf(x)$ , which one is yours? In general $operatorname{erfi}(x)=frac{2}{sqrt{pi}}int_0^xe^{z^2}dz,$ and $operatorname{erf}(x)=frac{2}{sqrt{pi}}int_0^xe^{-z^2}dz.$
$endgroup$
– Riemann
Sep 14 '18 at 9:25














$begingroup$
@Riemann. According to the result, it is $text{erfi}$
$endgroup$
– Claude Leibovici
Sep 14 '18 at 9:57




$begingroup$
@Riemann. According to the result, it is $text{erfi}$
$endgroup$
– Claude Leibovici
Sep 14 '18 at 9:57




1




1




$begingroup$
Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
$endgroup$
– Batominovski
Sep 14 '18 at 10:19






$begingroup$
Probably not useful, but $$I=int_0^infty,left(frac{sqrt{pi}}{2},text{erfc}(x),expleft(+x^2right)-frac{1}{1+2x}right),text{d}x,.$$ Here, $text{erfc}(x)=1-text{erf}(x)$ for all $xinmathbb{C}$. (Of course, I assume that $text{erf}$ and $text{erfi}$ are defined correctly as Riemann suggested.)
$endgroup$
– Batominovski
Sep 14 '18 at 10:19














$begingroup$
@KemonoChen Can you please clarify your question.
$endgroup$
– paulplusx
Sep 14 '18 at 13:16




$begingroup$
@KemonoChen Can you please clarify your question.
$endgroup$
– paulplusx
Sep 14 '18 at 13:16












$begingroup$
It is $operatorname{erfi}$. Sorry for the typo.
$endgroup$
– Kemono Chen
Sep 14 '18 at 13:16




$begingroup$
It is $operatorname{erfi}$. Sorry for the typo.
$endgroup$
– Kemono Chen
Sep 14 '18 at 13:16










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