Nice result that I can't prove: $int_{-2}^{2} tan^{-1} bigg( exp(-x²text{erf}(x)) bigg) ;dx=pi$












3














I'm always trying to find the integral representation of $pi$ using some interesting special function, at this time I have got the below representation
$$I=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx=pi$$ and according to Wolfram alpha its numerical value is very close to $pi$.



The problem that I have accrossed is the closed form of : $exp(-x^2text{erf}(x))$ in the range $[-2,2]$ , I have used $$mathrm{erf}!left(xright)^2approx1-expBig(-frac 4 {pi},frac{1+alpha x^2}{1+beta x^2},x^2 Big)$$ $$alpha=frac{10-pi ^2}{5 (pi -3) pi }$$ $$beta=frac{120-60 pi +7 pi ^2}{15 (pi -3) pi }.$$ The value of the corresponding error function is $1.1568times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $pi$? If no any simple way for integration ?






integration probability-distributions approximation pi error-function







share|cite|improve this question
























  • That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
    – AmbretteOrrisey
    Dec 3 '18 at 22:03
















3














I'm always trying to find the integral representation of $pi$ using some interesting special function, at this time I have got the below representation
$$I=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx=pi$$ and according to Wolfram alpha its numerical value is very close to $pi$.



The problem that I have accrossed is the closed form of : $exp(-x^2text{erf}(x))$ in the range $[-2,2]$ , I have used $$mathrm{erf}!left(xright)^2approx1-expBig(-frac 4 {pi},frac{1+alpha x^2}{1+beta x^2},x^2 Big)$$ $$alpha=frac{10-pi ^2}{5 (pi -3) pi }$$ $$beta=frac{120-60 pi +7 pi ^2}{15 (pi -3) pi }.$$ The value of the corresponding error function is $1.1568times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $pi$? If no any simple way for integration ?






integration probability-distributions approximation pi error-function







share|cite|improve this question
























  • That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
    – AmbretteOrrisey
    Dec 3 '18 at 22:03














3












3








3


1





I'm always trying to find the integral representation of $pi$ using some interesting special function, at this time I have got the below representation
$$I=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx=pi$$ and according to Wolfram alpha its numerical value is very close to $pi$.



The problem that I have accrossed is the closed form of : $exp(-x^2text{erf}(x))$ in the range $[-2,2]$ , I have used $$mathrm{erf}!left(xright)^2approx1-expBig(-frac 4 {pi},frac{1+alpha x^2}{1+beta x^2},x^2 Big)$$ $$alpha=frac{10-pi ^2}{5 (pi -3) pi }$$ $$beta=frac{120-60 pi +7 pi ^2}{15 (pi -3) pi }.$$ The value of the corresponding error function is $1.1568times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $pi$? If no any simple way for integration ?






integration probability-distributions approximation pi error-function







share|cite|improve this question















I'm always trying to find the integral representation of $pi$ using some interesting special function, at this time I have got the below representation
$$I=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx=pi$$ and according to Wolfram alpha its numerical value is very close to $pi$.



The problem that I have accrossed is the closed form of : $exp(-x^2text{erf}(x))$ in the range $[-2,2]$ , I have used $$mathrm{erf}!left(xright)^2approx1-expBig(-frac 4 {pi},frac{1+alpha x^2}{1+beta x^2},x^2 Big)$$ $$alpha=frac{10-pi ^2}{5 (pi -3) pi }$$ $$beta=frac{120-60 pi +7 pi ^2}{15 (pi -3) pi }.$$ The value of the corresponding error function is $1.1568times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $pi$? If no any simple way for integration ?






integration probability-distributions approximation pi error-function




integration probability-distributions approximation pi error-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 15:54









Robert Z

93.8k1061132




93.8k1061132










asked Dec 3 '18 at 19:34









zeraoulia rafikzeraoulia rafik

2,38711029




2,38711029












  • That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
    – AmbretteOrrisey
    Dec 3 '18 at 22:03


















  • That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
    – AmbretteOrrisey
    Dec 3 '18 at 22:03
















That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
– AmbretteOrrisey
Dec 3 '18 at 22:03




That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you?
– AmbretteOrrisey
Dec 3 '18 at 22:03










1 Answer
1






active

oldest

votes


















9














Yes, the integral $I$ is equal to $pi$. Note that for $t>0$
$$arctan(t)+arctan(1/t)=pi/2.$$
and after letting $y=-x$ we get
$$I:=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx
=int_{-2}^{2} tan^{-1} bigg( exp(y^2text{erf}(y)) bigg) ;dy.$$
Hence
$$I=frac{1}{2}int_{-2}^{2}arctan(t(x))+arctan(1/t(x))dx=frac{pi/2cdot 4}{2}=pi$$
where $t(x)=exp(-x^2text{erf}(x))$.



The same argument holds if we replace $-x^2text{erf}(x)$ with any odd function (see Michael Seifert's comment below).






share|cite|improve this answer



















  • 1




    Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
    – Michael Seifert
    Dec 3 '18 at 20:01












  • Attractive answer
    – zeraoulia rafik
    Dec 3 '18 at 20:07











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9














Yes, the integral $I$ is equal to $pi$. Note that for $t>0$
$$arctan(t)+arctan(1/t)=pi/2.$$
and after letting $y=-x$ we get
$$I:=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx
=int_{-2}^{2} tan^{-1} bigg( exp(y^2text{erf}(y)) bigg) ;dy.$$
Hence
$$I=frac{1}{2}int_{-2}^{2}arctan(t(x))+arctan(1/t(x))dx=frac{pi/2cdot 4}{2}=pi$$
where $t(x)=exp(-x^2text{erf}(x))$.



The same argument holds if we replace $-x^2text{erf}(x)$ with any odd function (see Michael Seifert's comment below).






share|cite|improve this answer



















  • 1




    Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
    – Michael Seifert
    Dec 3 '18 at 20:01












  • Attractive answer
    – zeraoulia rafik
    Dec 3 '18 at 20:07
















9














Yes, the integral $I$ is equal to $pi$. Note that for $t>0$
$$arctan(t)+arctan(1/t)=pi/2.$$
and after letting $y=-x$ we get
$$I:=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx
=int_{-2}^{2} tan^{-1} bigg( exp(y^2text{erf}(y)) bigg) ;dy.$$
Hence
$$I=frac{1}{2}int_{-2}^{2}arctan(t(x))+arctan(1/t(x))dx=frac{pi/2cdot 4}{2}=pi$$
where $t(x)=exp(-x^2text{erf}(x))$.



The same argument holds if we replace $-x^2text{erf}(x)$ with any odd function (see Michael Seifert's comment below).






share|cite|improve this answer



















  • 1




    Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
    – Michael Seifert
    Dec 3 '18 at 20:01












  • Attractive answer
    – zeraoulia rafik
    Dec 3 '18 at 20:07














9












9








9






Yes, the integral $I$ is equal to $pi$. Note that for $t>0$
$$arctan(t)+arctan(1/t)=pi/2.$$
and after letting $y=-x$ we get
$$I:=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx
=int_{-2}^{2} tan^{-1} bigg( exp(y^2text{erf}(y)) bigg) ;dy.$$
Hence
$$I=frac{1}{2}int_{-2}^{2}arctan(t(x))+arctan(1/t(x))dx=frac{pi/2cdot 4}{2}=pi$$
where $t(x)=exp(-x^2text{erf}(x))$.



The same argument holds if we replace $-x^2text{erf}(x)$ with any odd function (see Michael Seifert's comment below).






share|cite|improve this answer














Yes, the integral $I$ is equal to $pi$. Note that for $t>0$
$$arctan(t)+arctan(1/t)=pi/2.$$
and after letting $y=-x$ we get
$$I:=int_{-2}^{2} tan^{-1} bigg( exp(-x^2text{erf}(x)) bigg) ;dx
=int_{-2}^{2} tan^{-1} bigg( exp(y^2text{erf}(y)) bigg) ;dy.$$
Hence
$$I=frac{1}{2}int_{-2}^{2}arctan(t(x))+arctan(1/t(x))dx=frac{pi/2cdot 4}{2}=pi$$
where $t(x)=exp(-x^2text{erf}(x))$.



The same argument holds if we replace $-x^2text{erf}(x)$ with any odd function (see Michael Seifert's comment below).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 5:20

























answered Dec 3 '18 at 19:55









Robert ZRobert Z

93.8k1061132




93.8k1061132








  • 1




    Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
    – Michael Seifert
    Dec 3 '18 at 20:01












  • Attractive answer
    – zeraoulia rafik
    Dec 3 '18 at 20:07














  • 1




    Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
    – Michael Seifert
    Dec 3 '18 at 20:01












  • Attractive answer
    – zeraoulia rafik
    Dec 3 '18 at 20:07








1




1




Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
– Michael Seifert
Dec 3 '18 at 20:01






Note that this argument would still work, and the integral would still be $pi$, if you replaced $x^2 mathrm{erf}(x)$ by any odd function $f(x)$.
– Michael Seifert
Dec 3 '18 at 20:01














Attractive answer
– zeraoulia rafik
Dec 3 '18 at 20:07




Attractive answer
– zeraoulia rafik
Dec 3 '18 at 20:07


















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