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How calculate $quad int_0^infty fraccos(x^2)1+x^2 dx$
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How calculate
$$int_0^infty fraccos(x^2)1+x^2;dx$$
$mathbf My Attempt$
I tried introducing a new parameter and differentiating twice like this:
$$I(a)=int_0^infty fraccos(ax^2)1+x^2;dx quad Rightarrow$$
$$I^''(a)+I(a)=fracsqrtpisqrt2a^frac-12+fracsqrtpi2sqrt2a^frac-32$$
I'm unable to come up with a particular solution for this differential equation.
I tried using a double integral like this:
$$int_0^infty e^-y(1+x^2); dy = frac11+x^2 quadRightarrow$$
$$int_0^infty int_0^infty cos(x^2); e^-y(1+x^2); dy,dx = int_0^inftyfraccos(x^2)1+x^2; dx$$
But this wasn't that useful.
Result by Wolfram (Mathematica): $$-frac12 pi left[sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right] $$
Any hint? (I'm not familiar with the Residue Theorem)
improper-integrals
asked Jul 20 at 4:57
Wolfdale
24919
 |Â
show 1 more comment
up vote
3
down vote
favorite
How calculate
$$int_0^infty fraccos(x^2)1+x^2;dx$$
$mathbf My Attempt$
I tried introducing a new parameter and differentiating twice like this:
$$I(a)=int_0^infty fraccos(ax^2)1+x^2;dx quad Rightarrow$$
$$I^''(a)+I(a)=fracsqrtpisqrt2a^frac-12+fracsqrtpi2sqrt2a^frac-32$$
I'm unable to come up with a particular solution for this differential equation.
I tried using a double integral like this:
$$int_0^infty e^-y(1+x^2); dy = frac11+x^2 quadRightarrow$$
$$int_0^infty int_0^infty cos(x^2); e^-y(1+x^2); dy,dx = int_0^inftyfraccos(x^2)1+x^2; dx$$
But this wasn't that useful.
Result by Wolfram (Mathematica): $$-frac12 pi left[sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right] $$
Any hint? (I'm not familiar with the Residue Theorem)
improper-integrals
asked Jul 20 at 4:57
Wolfdale
24919
Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How calculate
$$int_0^infty fraccos(x^2)1+x^2;dx$$
$mathbf My Attempt$
I tried introducing a new parameter and differentiating twice like this:
$$I(a)=int_0^infty fraccos(ax^2)1+x^2;dx quad Rightarrow$$
$$I^''(a)+I(a)=fracsqrtpisqrt2a^frac-12+fracsqrtpi2sqrt2a^frac-32$$
I'm unable to come up with a particular solution for this differential equation.
I tried using a double integral like this:
$$int_0^infty e^-y(1+x^2); dy = frac11+x^2 quadRightarrow$$
$$int_0^infty int_0^infty cos(x^2); e^-y(1+x^2); dy,dx = int_0^inftyfraccos(x^2)1+x^2; dx$$
But this wasn't that useful.
Result by Wolfram (Mathematica): $$-frac12 pi left[sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right] $$
Any hint? (I'm not familiar with the Residue Theorem)
improper-integrals
asked Jul 20 at 4:57
Wolfdale
24919
How calculate
$$int_0^infty fraccos(x^2)1+x^2;dx$$
$mathbf My Attempt$
I tried introducing a new parameter and differentiating twice like this:
$$I(a)=int_0^infty fraccos(ax^2)1+x^2;dx quad Rightarrow$$
$$I^''(a)+I(a)=fracsqrtpisqrt2a^frac-12+fracsqrtpi2sqrt2a^frac-32$$
I'm unable to come up with a particular solution for this differential equation.
I tried using a double integral like this:
$$int_0^infty e^-y(1+x^2); dy = frac11+x^2 quadRightarrow$$
$$int_0^infty int_0^infty cos(x^2); e^-y(1+x^2); dy,dx = int_0^inftyfraccos(x^2)1+x^2; dx$$
But this wasn't that useful.
Result by Wolfram (Mathematica): $$-frac12 pi left[sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right] $$
Any hint? (I'm not familiar with the Residue Theorem)
improper-integrals
asked Jul 20 at 4:57
Wolfdale
24919
edited Jul 20 at 6:18
asked Jul 20 at 4:57
Wolfdale
24919
asked Jul 20 at 4:57
Wolfdale
24919
asked Jul 20 at 4:57
Wolfdale
24919
24919
Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14
 |Â
show 1 more comment
Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14
Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=int_0^inftyfrace^iax^2x^2+1dx,$$
$$I'(a)=iint_0^infty x^2frace^iax^2x^2+1dx,$$
$$I'(a)+iI(a)=iint_0^infty e^iax^2dx=wa^-1/2,$$ where $w$ is a complex constant (namely $(i-1)sqrtpi/8$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^ia=(I(a)e^ia)'=wa^-1/2e^ia$$
and integrating from $a=0$ to $1$,
$$I(a)e^ia-I(0)=wint_0^1a^-1/2e^iada=2wint_0^1e^ib^2db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=left(2w(C(1)+iS(1))+I(0)right)e^-i$$ of which you take the real part. (With $I(0)=pi/2$.)
Note that we have been using the Fresnel integral without the $pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
answered Jul 20 at 7:18
Yves Daoust
111k665204
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]]gives the correct result
– robjohn♦
Jul 21 at 16:56
add a comment |Â
up vote
1
down vote
With a bit of contour integration, we can speed up the convergence:
$$newcommandReoperatornameRe
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(int_0^inftyfrace^ix^21+x^2,mathrmdxright)tag1\
&=Releft(frac1+isqrt2int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag2\
&=frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdxtag3
endalign
$$
Explanation:
$(1)$: $cosleft(x^2right)=Releft(e^ix^2right)$
$(2)$: substitute $xmapstofrac1+isqrt2x$
$phantom(2)text:$ change contour from $frac1-isqrt2[0,infty]$ to $[0,infty]$ using Cauchy Integral Theorem
$(3)$: extract real part
Numerically integrating $(3)$ gives
$$
frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdx
=0.65280425451173388408tag4
$$
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(e^ipi/4int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag5\
&=Releft(e^i3pi/4e^-iint_0^inftyfrace^i-x^2i-x^2,mathrmdxright)tag6\
endalign
$$
Explanation:
$(5)$: copy $(2)$
$(6)$: multiply by $1=left(e^ipi/2e^-iright)left(e^i/iright)$
Define
$$
f(a)=Releft(e^i3pi/4e^-iint_0^inftyfrace^aleft(i-x^2right)i-x^2,mathrmdxright)tag7
$$
Then
$$newcommandResoperatorname*Res
beginalign
f(0)
&=Releft(e^i3pi/4e^-iint_0^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ifrac12int_-infty^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ipi iRes_z=e^ipi/4left(frac1i-z^2right)right)\
&=Releft(fracpi2e^-iright)tag8
endalign
$$
Taking the derivative of $(7)$, we have
$$
beginalign
f'(a)
&=Releft(e^i3pi/4e^-iint_0^infty e^aleft(i-x^2right),mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iaint_0^infty e^-ax^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iasqrtfracpi4aright)tag9
endalign
$$
Integrating $(9)$ gives
$$newcommanderfoperatornameerfnewcommanderfcoperatornameerfc
beginalign
f(1)-f(0)
&=Releft(fracsqrtpi2e^i3pi/4e^-iint_0^1frace^iasqrta,mathrmdaright)tag10\
&=Releft(sqrtpi e^i3pi/4e^-iint_0^1e^ia^2,mathrmdaright)tag11\
&=Releft(sqrtpi e^ipie^-iint_0^e^-ipi/4e^-a^2,mathrmdaright)tag12\[3pt]
&=Releft(-fracpi2e^-ioperatornameerfleft(e^-ipi/4right)right)tag13
endalign
$$
Explanation:
$(10)$: integrate $(9)$ from $0$ to $1$
$(11)$: substitute $amapsto a^2$
$(12)$: substitute $amapsto ae^ipi/4$
$(13)$: use $erf$
Combining $(6)$, $(7)$, $(8)$, $(13)$, and $erfc(x)=1-erf(x)$, yields
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(fracpi2e^-ierfcleft(e^-ipi/4right)right)\
&=0.65280425451173388408tag14
endalign
$$
Thankfully, $(4)$ and $(14)$ agree.
answered Jul 21 at 13:57
robjohn♦
258k26297612
add a comment |Â
up vote
1
down vote
$$J=int^infty_0fraccos(x^2)1+x^2dx=Reint^infty_0frace^ix^21+x^2dx=Re,, e^-iunderbraceint^infty_0frace^ia(x^2+1)1+x^2dx_=I(a)$$ where $a=1$.
Restrict $a$ to be real non-negative.
$$I'(a)=ie^iaint^infty_0e^iax^2dx$$
By the substitution $-u=iax^2$, we get
$$I'(a)=frac12ie^iasqrtfraciaint^-iinfty_0u^-1/2e^-udu=fraci^3/22frace^iasqrt asqrtpi$$
Let $k=fraci^3/22sqrtpi$.
$$I(a)=kintfrace^iasqrt ada$$
By the substitution $ia=-v$, we get
$$I(a)=ksqrt iintfrace^-vsqrt vdv=-fracsqrtpi2gamma(frac12,v)+C=-fracsqrtpi2gamma(frac12,-ia)+C$$
Because $I(0)=fracpi2$, so $C=fracpi2$.
Since $gamma(frac12,x)=sqrtpitexterf(sqrt x)$,
$$I(a)=-fracpi2texterf(sqrt -isqrt a)+fracpi2=fracpi2texterfc(overbracee^-pi i/4^textprincipal valuecdotsqrt a)$$
As a result, $$J=Re e^-iI(1)=fracpi2,Re,, e^-itexterfc(e^-pi i/4)$$
Note that $I(1)=-fracsqrtpi2gamma(frac12,-i)+fracpi2$ is wrong if you calculate it using Wolfram Alpha. However, an equivalent expression $I(1) = fracpi2texterfc(e^-pi i/4)$ is correct as calculated by Wolfram Alpha. I don't know why $-fracsqrtpi2gamma(frac12,-i)+fracpi2ne fracpi2texterfc(e^-pi i/4)$. I suspect this might be a bug of Wolfy.
answered Jul 20 at 7:19
Szeto
4,1731521
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I])gives the same numerical result asPi/2Erf[Exp[-I Pi/4]]
– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=int_0^inftyfrace^iax^2x^2+1dx,$$
$$I'(a)=iint_0^infty x^2frace^iax^2x^2+1dx,$$
$$I'(a)+iI(a)=iint_0^infty e^iax^2dx=wa^-1/2,$$ where $w$ is a complex constant (namely $(i-1)sqrtpi/8$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^ia=(I(a)e^ia)'=wa^-1/2e^ia$$
and integrating from $a=0$ to $1$,
$$I(a)e^ia-I(0)=wint_0^1a^-1/2e^iada=2wint_0^1e^ib^2db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=left(2w(C(1)+iS(1))+I(0)right)e^-i$$ of which you take the real part. (With $I(0)=pi/2$.)
Note that we have been using the Fresnel integral without the $pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
answered Jul 20 at 7:18
Yves Daoust
111k665204
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]]gives the correct result
– robjohn♦
Jul 21 at 16:56
add a comment |Â
up vote
2
down vote
accepted
Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=int_0^inftyfrace^iax^2x^2+1dx,$$
$$I'(a)=iint_0^infty x^2frace^iax^2x^2+1dx,$$
$$I'(a)+iI(a)=iint_0^infty e^iax^2dx=wa^-1/2,$$ where $w$ is a complex constant (namely $(i-1)sqrtpi/8$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^ia=(I(a)e^ia)'=wa^-1/2e^ia$$
and integrating from $a=0$ to $1$,
$$I(a)e^ia-I(0)=wint_0^1a^-1/2e^iada=2wint_0^1e^ib^2db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=left(2w(C(1)+iS(1))+I(0)right)e^-i$$ of which you take the real part. (With $I(0)=pi/2$.)
Note that we have been using the Fresnel integral without the $pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
answered Jul 20 at 7:18
Yves Daoust
111k665204
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]]gives the correct result
– robjohn♦
Jul 21 at 16:56
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=int_0^inftyfrace^iax^2x^2+1dx,$$
$$I'(a)=iint_0^infty x^2frace^iax^2x^2+1dx,$$
$$I'(a)+iI(a)=iint_0^infty e^iax^2dx=wa^-1/2,$$ where $w$ is a complex constant (namely $(i-1)sqrtpi/8$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^ia=(I(a)e^ia)'=wa^-1/2e^ia$$
and integrating from $a=0$ to $1$,
$$I(a)e^ia-I(0)=wint_0^1a^-1/2e^iada=2wint_0^1e^ib^2db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=left(2w(C(1)+iS(1))+I(0)right)e^-i$$ of which you take the real part. (With $I(0)=pi/2$.)
Note that we have been using the Fresnel integral without the $pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
answered Jul 20 at 7:18
Yves Daoust
111k665204
Using the complex representation makes the problem solvable with a first order ODE.
$$I(a)=int_0^inftyfrace^iax^2x^2+1dx,$$
$$I'(a)=iint_0^infty x^2frace^iax^2x^2+1dx,$$
$$I'(a)+iI(a)=iint_0^infty e^iax^2dx=wa^-1/2,$$ where $w$ is a complex constant (namely $(i-1)sqrtpi/8$).
Then by means of an integrating factor,
$$(I'(a)+iI(a))e^ia=(I(a)e^ia)'=wa^-1/2e^ia$$
and integrating from $a=0$ to $1$,
$$I(a)e^ia-I(0)=wint_0^1a^-1/2e^iada=2wint_0^1e^ib^2db=2w(C(1)+iS(1)).$$
Finally,
$$I(a)=left(2w(C(1)+iS(1))+I(0)right)e^-i$$ of which you take the real part. (With $I(0)=pi/2$.)
Note that we have been using the Fresnel integral without the $pi/2$ factor in its definition, and this answer coincides with those of the CAS softwares.
answered Jul 20 at 7:18
Yves Daoust
111k665204
edited Jul 20 at 7:30
answered Jul 20 at 7:18
Yves Daoust
111k665204
answered Jul 20 at 7:18
Yves Daoust
111k665204
answered Jul 20 at 7:18
Yves Daoust
111k665204
111k665204
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]]gives the correct result
– robjohn♦
Jul 21 at 16:56
add a comment |Â
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]]gives the correct result
– robjohn♦
Jul 21 at 16:56
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
Brilliant, actually I went the same way but found the calculations complicated and stopped. Many thx.
– Wolfdale
Jul 20 at 17:24
In Mathematica,
Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]] gives the correct result– robjohn♦
Jul 21 at 16:56
In Mathematica,
Re[((I-1)Sqrt[Pi/2](Sqrt[Pi/2]FresnelC[Sqrt[2/Pi]] + I Sqrt[Pi/2]FresnelS[Sqrt[2/Pi]])+Pi/2)Exp[-I]] gives the correct result– robjohn♦
Jul 21 at 16:56
add a comment |Â
up vote
1
down vote
With a bit of contour integration, we can speed up the convergence:
$$newcommandReoperatornameRe
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(int_0^inftyfrace^ix^21+x^2,mathrmdxright)tag1\
&=Releft(frac1+isqrt2int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag2\
&=frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdxtag3
endalign
$$
Explanation:
$(1)$: $cosleft(x^2right)=Releft(e^ix^2right)$
$(2)$: substitute $xmapstofrac1+isqrt2x$
$phantom(2)text:$ change contour from $frac1-isqrt2[0,infty]$ to $[0,infty]$ using Cauchy Integral Theorem
$(3)$: extract real part
Numerically integrating $(3)$ gives
$$
frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdx
=0.65280425451173388408tag4
$$
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(e^ipi/4int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag5\
&=Releft(e^i3pi/4e^-iint_0^inftyfrace^i-x^2i-x^2,mathrmdxright)tag6\
endalign
$$
Explanation:
$(5)$: copy $(2)$
$(6)$: multiply by $1=left(e^ipi/2e^-iright)left(e^i/iright)$
Define
$$
f(a)=Releft(e^i3pi/4e^-iint_0^inftyfrace^aleft(i-x^2right)i-x^2,mathrmdxright)tag7
$$
Then
$$newcommandResoperatorname*Res
beginalign
f(0)
&=Releft(e^i3pi/4e^-iint_0^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ifrac12int_-infty^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ipi iRes_z=e^ipi/4left(frac1i-z^2right)right)\
&=Releft(fracpi2e^-iright)tag8
endalign
$$
Taking the derivative of $(7)$, we have
$$
beginalign
f'(a)
&=Releft(e^i3pi/4e^-iint_0^infty e^aleft(i-x^2right),mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iaint_0^infty e^-ax^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iasqrtfracpi4aright)tag9
endalign
$$
Integrating $(9)$ gives
$$newcommanderfoperatornameerfnewcommanderfcoperatornameerfc
beginalign
f(1)-f(0)
&=Releft(fracsqrtpi2e^i3pi/4e^-iint_0^1frace^iasqrta,mathrmdaright)tag10\
&=Releft(sqrtpi e^i3pi/4e^-iint_0^1e^ia^2,mathrmdaright)tag11\
&=Releft(sqrtpi e^ipie^-iint_0^e^-ipi/4e^-a^2,mathrmdaright)tag12\[3pt]
&=Releft(-fracpi2e^-ioperatornameerfleft(e^-ipi/4right)right)tag13
endalign
$$
Explanation:
$(10)$: integrate $(9)$ from $0$ to $1$
$(11)$: substitute $amapsto a^2$
$(12)$: substitute $amapsto ae^ipi/4$
$(13)$: use $erf$
Combining $(6)$, $(7)$, $(8)$, $(13)$, and $erfc(x)=1-erf(x)$, yields
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(fracpi2e^-ierfcleft(e^-ipi/4right)right)\
&=0.65280425451173388408tag14
endalign
$$
Thankfully, $(4)$ and $(14)$ agree.
answered Jul 21 at 13:57
robjohn♦
258k26297612
add a comment |Â
up vote
1
down vote
With a bit of contour integration, we can speed up the convergence:
$$newcommandReoperatornameRe
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(int_0^inftyfrace^ix^21+x^2,mathrmdxright)tag1\
&=Releft(frac1+isqrt2int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag2\
&=frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdxtag3
endalign
$$
Explanation:
$(1)$: $cosleft(x^2right)=Releft(e^ix^2right)$
$(2)$: substitute $xmapstofrac1+isqrt2x$
$phantom(2)text:$ change contour from $frac1-isqrt2[0,infty]$ to $[0,infty]$ using Cauchy Integral Theorem
$(3)$: extract real part
Numerically integrating $(3)$ gives
$$
frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdx
=0.65280425451173388408tag4
$$
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(e^ipi/4int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag5\
&=Releft(e^i3pi/4e^-iint_0^inftyfrace^i-x^2i-x^2,mathrmdxright)tag6\
endalign
$$
Explanation:
$(5)$: copy $(2)$
$(6)$: multiply by $1=left(e^ipi/2e^-iright)left(e^i/iright)$
Define
$$
f(a)=Releft(e^i3pi/4e^-iint_0^inftyfrace^aleft(i-x^2right)i-x^2,mathrmdxright)tag7
$$
Then
$$newcommandResoperatorname*Res
beginalign
f(0)
&=Releft(e^i3pi/4e^-iint_0^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ifrac12int_-infty^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ipi iRes_z=e^ipi/4left(frac1i-z^2right)right)\
&=Releft(fracpi2e^-iright)tag8
endalign
$$
Taking the derivative of $(7)$, we have
$$
beginalign
f'(a)
&=Releft(e^i3pi/4e^-iint_0^infty e^aleft(i-x^2right),mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iaint_0^infty e^-ax^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iasqrtfracpi4aright)tag9
endalign
$$
Integrating $(9)$ gives
$$newcommanderfoperatornameerfnewcommanderfcoperatornameerfc
beginalign
f(1)-f(0)
&=Releft(fracsqrtpi2e^i3pi/4e^-iint_0^1frace^iasqrta,mathrmdaright)tag10\
&=Releft(sqrtpi e^i3pi/4e^-iint_0^1e^ia^2,mathrmdaright)tag11\
&=Releft(sqrtpi e^ipie^-iint_0^e^-ipi/4e^-a^2,mathrmdaright)tag12\[3pt]
&=Releft(-fracpi2e^-ioperatornameerfleft(e^-ipi/4right)right)tag13
endalign
$$
Explanation:
$(10)$: integrate $(9)$ from $0$ to $1$
$(11)$: substitute $amapsto a^2$
$(12)$: substitute $amapsto ae^ipi/4$
$(13)$: use $erf$
Combining $(6)$, $(7)$, $(8)$, $(13)$, and $erfc(x)=1-erf(x)$, yields
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(fracpi2e^-ierfcleft(e^-ipi/4right)right)\
&=0.65280425451173388408tag14
endalign
$$
Thankfully, $(4)$ and $(14)$ agree.
answered Jul 21 at 13:57
robjohn♦
258k26297612
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With a bit of contour integration, we can speed up the convergence:
$$newcommandReoperatornameRe
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(int_0^inftyfrace^ix^21+x^2,mathrmdxright)tag1\
&=Releft(frac1+isqrt2int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag2\
&=frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdxtag3
endalign
$$
Explanation:
$(1)$: $cosleft(x^2right)=Releft(e^ix^2right)$
$(2)$: substitute $xmapstofrac1+isqrt2x$
$phantom(2)text:$ change contour from $frac1-isqrt2[0,infty]$ to $[0,infty]$ using Cauchy Integral Theorem
$(3)$: extract real part
Numerically integrating $(3)$ gives
$$
frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdx
=0.65280425451173388408tag4
$$
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(e^ipi/4int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag5\
&=Releft(e^i3pi/4e^-iint_0^inftyfrace^i-x^2i-x^2,mathrmdxright)tag6\
endalign
$$
Explanation:
$(5)$: copy $(2)$
$(6)$: multiply by $1=left(e^ipi/2e^-iright)left(e^i/iright)$
Define
$$
f(a)=Releft(e^i3pi/4e^-iint_0^inftyfrace^aleft(i-x^2right)i-x^2,mathrmdxright)tag7
$$
Then
$$newcommandResoperatorname*Res
beginalign
f(0)
&=Releft(e^i3pi/4e^-iint_0^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ifrac12int_-infty^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ipi iRes_z=e^ipi/4left(frac1i-z^2right)right)\
&=Releft(fracpi2e^-iright)tag8
endalign
$$
Taking the derivative of $(7)$, we have
$$
beginalign
f'(a)
&=Releft(e^i3pi/4e^-iint_0^infty e^aleft(i-x^2right),mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iaint_0^infty e^-ax^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iasqrtfracpi4aright)tag9
endalign
$$
Integrating $(9)$ gives
$$newcommanderfoperatornameerfnewcommanderfcoperatornameerfc
beginalign
f(1)-f(0)
&=Releft(fracsqrtpi2e^i3pi/4e^-iint_0^1frace^iasqrta,mathrmdaright)tag10\
&=Releft(sqrtpi e^i3pi/4e^-iint_0^1e^ia^2,mathrmdaright)tag11\
&=Releft(sqrtpi e^ipie^-iint_0^e^-ipi/4e^-a^2,mathrmdaright)tag12\[3pt]
&=Releft(-fracpi2e^-ioperatornameerfleft(e^-ipi/4right)right)tag13
endalign
$$
Explanation:
$(10)$: integrate $(9)$ from $0$ to $1$
$(11)$: substitute $amapsto a^2$
$(12)$: substitute $amapsto ae^ipi/4$
$(13)$: use $erf$
Combining $(6)$, $(7)$, $(8)$, $(13)$, and $erfc(x)=1-erf(x)$, yields
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(fracpi2e^-ierfcleft(e^-ipi/4right)right)\
&=0.65280425451173388408tag14
endalign
$$
Thankfully, $(4)$ and $(14)$ agree.
answered Jul 21 at 13:57
robjohn♦
258k26297612
With a bit of contour integration, we can speed up the convergence:
$$newcommandReoperatornameRe
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(int_0^inftyfrace^ix^21+x^2,mathrmdxright)tag1\
&=Releft(frac1+isqrt2int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag2\
&=frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdxtag3
endalign
$$
Explanation:
$(1)$: $cosleft(x^2right)=Releft(e^ix^2right)$
$(2)$: substitute $xmapstofrac1+isqrt2x$
$phantom(2)text:$ change contour from $frac1-isqrt2[0,infty]$ to $[0,infty]$ using Cauchy Integral Theorem
$(3)$: extract real part
Numerically integrating $(3)$ gives
$$
frac1sqrt2int_0^inftyfrac1+x^21+x^4e^-x^2,mathrmdx
=0.65280425451173388408tag4
$$
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(e^ipi/4int_0^inftyfrace^-x^21+ix^2,mathrmdxright)tag5\
&=Releft(e^i3pi/4e^-iint_0^inftyfrace^i-x^2i-x^2,mathrmdxright)tag6\
endalign
$$
Explanation:
$(5)$: copy $(2)$
$(6)$: multiply by $1=left(e^ipi/2e^-iright)left(e^i/iright)$
Define
$$
f(a)=Releft(e^i3pi/4e^-iint_0^inftyfrace^aleft(i-x^2right)i-x^2,mathrmdxright)tag7
$$
Then
$$newcommandResoperatorname*Res
beginalign
f(0)
&=Releft(e^i3pi/4e^-iint_0^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ifrac12int_-infty^inftyfrac1i-x^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ipi iRes_z=e^ipi/4left(frac1i-z^2right)right)\
&=Releft(fracpi2e^-iright)tag8
endalign
$$
Taking the derivative of $(7)$, we have
$$
beginalign
f'(a)
&=Releft(e^i3pi/4e^-iint_0^infty e^aleft(i-x^2right),mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iaint_0^infty e^-ax^2,mathrmdxright)\
&=Releft(e^i3pi/4e^-ie^iasqrtfracpi4aright)tag9
endalign
$$
Integrating $(9)$ gives
$$newcommanderfoperatornameerfnewcommanderfcoperatornameerfc
beginalign
f(1)-f(0)
&=Releft(fracsqrtpi2e^i3pi/4e^-iint_0^1frace^iasqrta,mathrmdaright)tag10\
&=Releft(sqrtpi e^i3pi/4e^-iint_0^1e^ia^2,mathrmdaright)tag11\
&=Releft(sqrtpi e^ipie^-iint_0^e^-ipi/4e^-a^2,mathrmdaright)tag12\[3pt]
&=Releft(-fracpi2e^-ioperatornameerfleft(e^-ipi/4right)right)tag13
endalign
$$
Explanation:
$(10)$: integrate $(9)$ from $0$ to $1$
$(11)$: substitute $amapsto a^2$
$(12)$: substitute $amapsto ae^ipi/4$
$(13)$: use $erf$
Combining $(6)$, $(7)$, $(8)$, $(13)$, and $erfc(x)=1-erf(x)$, yields
$$
beginalign
int_0^inftyfraccosleft(x^2right)1+x^2,mathrmdx
&=Releft(fracpi2e^-ierfcleft(e^-ipi/4right)right)\
&=0.65280425451173388408tag14
endalign
$$
Thankfully, $(4)$ and $(14)$ agree.
answered Jul 21 at 13:57
robjohn♦
258k26297612
answered Jul 21 at 13:57
robjohn♦
258k26297612
answered Jul 21 at 13:57
robjohn♦
258k26297612
answered Jul 21 at 13:57
robjohn♦
258k26297612
258k26297612
add a comment |Â
add a comment |Â
up vote
1
down vote
$$J=int^infty_0fraccos(x^2)1+x^2dx=Reint^infty_0frace^ix^21+x^2dx=Re,, e^-iunderbraceint^infty_0frace^ia(x^2+1)1+x^2dx_=I(a)$$ where $a=1$.
Restrict $a$ to be real non-negative.
$$I'(a)=ie^iaint^infty_0e^iax^2dx$$
By the substitution $-u=iax^2$, we get
$$I'(a)=frac12ie^iasqrtfraciaint^-iinfty_0u^-1/2e^-udu=fraci^3/22frace^iasqrt asqrtpi$$
Let $k=fraci^3/22sqrtpi$.
$$I(a)=kintfrace^iasqrt ada$$
By the substitution $ia=-v$, we get
$$I(a)=ksqrt iintfrace^-vsqrt vdv=-fracsqrtpi2gamma(frac12,v)+C=-fracsqrtpi2gamma(frac12,-ia)+C$$
Because $I(0)=fracpi2$, so $C=fracpi2$.
Since $gamma(frac12,x)=sqrtpitexterf(sqrt x)$,
$$I(a)=-fracpi2texterf(sqrt -isqrt a)+fracpi2=fracpi2texterfc(overbracee^-pi i/4^textprincipal valuecdotsqrt a)$$
As a result, $$J=Re e^-iI(1)=fracpi2,Re,, e^-itexterfc(e^-pi i/4)$$
Note that $I(1)=-fracsqrtpi2gamma(frac12,-i)+fracpi2$ is wrong if you calculate it using Wolfram Alpha. However, an equivalent expression $I(1) = fracpi2texterfc(e^-pi i/4)$ is correct as calculated by Wolfram Alpha. I don't know why $-fracsqrtpi2gamma(frac12,-i)+fracpi2ne fracpi2texterfc(e^-pi i/4)$. I suspect this might be a bug of Wolfy.
answered Jul 20 at 7:19
Szeto
4,1731521
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I])gives the same numerical result asPi/2Erf[Exp[-I Pi/4]]
– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
 |Â
show 3 more comments
up vote
1
down vote
$$J=int^infty_0fraccos(x^2)1+x^2dx=Reint^infty_0frace^ix^21+x^2dx=Re,, e^-iunderbraceint^infty_0frace^ia(x^2+1)1+x^2dx_=I(a)$$ where $a=1$.
Restrict $a$ to be real non-negative.
$$I'(a)=ie^iaint^infty_0e^iax^2dx$$
By the substitution $-u=iax^2$, we get
$$I'(a)=frac12ie^iasqrtfraciaint^-iinfty_0u^-1/2e^-udu=fraci^3/22frace^iasqrt asqrtpi$$
Let $k=fraci^3/22sqrtpi$.
$$I(a)=kintfrace^iasqrt ada$$
By the substitution $ia=-v$, we get
$$I(a)=ksqrt iintfrace^-vsqrt vdv=-fracsqrtpi2gamma(frac12,v)+C=-fracsqrtpi2gamma(frac12,-ia)+C$$
Because $I(0)=fracpi2$, so $C=fracpi2$.
Since $gamma(frac12,x)=sqrtpitexterf(sqrt x)$,
$$I(a)=-fracpi2texterf(sqrt -isqrt a)+fracpi2=fracpi2texterfc(overbracee^-pi i/4^textprincipal valuecdotsqrt a)$$
As a result, $$J=Re e^-iI(1)=fracpi2,Re,, e^-itexterfc(e^-pi i/4)$$
Note that $I(1)=-fracsqrtpi2gamma(frac12,-i)+fracpi2$ is wrong if you calculate it using Wolfram Alpha. However, an equivalent expression $I(1) = fracpi2texterfc(e^-pi i/4)$ is correct as calculated by Wolfram Alpha. I don't know why $-fracsqrtpi2gamma(frac12,-i)+fracpi2ne fracpi2texterfc(e^-pi i/4)$. I suspect this might be a bug of Wolfy.
answered Jul 20 at 7:19
Szeto
4,1731521
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I])gives the same numerical result asPi/2Erf[Exp[-I Pi/4]]
– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
$$J=int^infty_0fraccos(x^2)1+x^2dx=Reint^infty_0frace^ix^21+x^2dx=Re,, e^-iunderbraceint^infty_0frace^ia(x^2+1)1+x^2dx_=I(a)$$ where $a=1$.
Restrict $a$ to be real non-negative.
$$I'(a)=ie^iaint^infty_0e^iax^2dx$$
By the substitution $-u=iax^2$, we get
$$I'(a)=frac12ie^iasqrtfraciaint^-iinfty_0u^-1/2e^-udu=fraci^3/22frace^iasqrt asqrtpi$$
Let $k=fraci^3/22sqrtpi$.
$$I(a)=kintfrace^iasqrt ada$$
By the substitution $ia=-v$, we get
$$I(a)=ksqrt iintfrace^-vsqrt vdv=-fracsqrtpi2gamma(frac12,v)+C=-fracsqrtpi2gamma(frac12,-ia)+C$$
Because $I(0)=fracpi2$, so $C=fracpi2$.
Since $gamma(frac12,x)=sqrtpitexterf(sqrt x)$,
$$I(a)=-fracpi2texterf(sqrt -isqrt a)+fracpi2=fracpi2texterfc(overbracee^-pi i/4^textprincipal valuecdotsqrt a)$$
As a result, $$J=Re e^-iI(1)=fracpi2,Re,, e^-itexterfc(e^-pi i/4)$$
Note that $I(1)=-fracsqrtpi2gamma(frac12,-i)+fracpi2$ is wrong if you calculate it using Wolfram Alpha. However, an equivalent expression $I(1) = fracpi2texterfc(e^-pi i/4)$ is correct as calculated by Wolfram Alpha. I don't know why $-fracsqrtpi2gamma(frac12,-i)+fracpi2ne fracpi2texterfc(e^-pi i/4)$. I suspect this might be a bug of Wolfy.
answered Jul 20 at 7:19
Szeto
4,1731521
$$J=int^infty_0fraccos(x^2)1+x^2dx=Reint^infty_0frace^ix^21+x^2dx=Re,, e^-iunderbraceint^infty_0frace^ia(x^2+1)1+x^2dx_=I(a)$$ where $a=1$.
Restrict $a$ to be real non-negative.
$$I'(a)=ie^iaint^infty_0e^iax^2dx$$
By the substitution $-u=iax^2$, we get
$$I'(a)=frac12ie^iasqrtfraciaint^-iinfty_0u^-1/2e^-udu=fraci^3/22frace^iasqrt asqrtpi$$
Let $k=fraci^3/22sqrtpi$.
$$I(a)=kintfrace^iasqrt ada$$
By the substitution $ia=-v$, we get
$$I(a)=ksqrt iintfrace^-vsqrt vdv=-fracsqrtpi2gamma(frac12,v)+C=-fracsqrtpi2gamma(frac12,-ia)+C$$
Because $I(0)=fracpi2$, so $C=fracpi2$.
Since $gamma(frac12,x)=sqrtpitexterf(sqrt x)$,
$$I(a)=-fracpi2texterf(sqrt -isqrt a)+fracpi2=fracpi2texterfc(overbracee^-pi i/4^textprincipal valuecdotsqrt a)$$
As a result, $$J=Re e^-iI(1)=fracpi2,Re,, e^-itexterfc(e^-pi i/4)$$
Note that $I(1)=-fracsqrtpi2gamma(frac12,-i)+fracpi2$ is wrong if you calculate it using Wolfram Alpha. However, an equivalent expression $I(1) = fracpi2texterfc(e^-pi i/4)$ is correct as calculated by Wolfram Alpha. I don't know why $-fracsqrtpi2gamma(frac12,-i)+fracpi2ne fracpi2texterfc(e^-pi i/4)$. I suspect this might be a bug of Wolfy.
answered Jul 20 at 7:19
Szeto
4,1731521
edited Jul 21 at 15:02
answered Jul 20 at 7:19
Szeto
4,1731521
answered Jul 20 at 7:19
Szeto
4,1731521
answered Jul 20 at 7:19
Szeto
4,1731521
4,1731521
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I])gives the same numerical result asPi/2Erf[Exp[-I Pi/4]]
– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
 |Â
show 3 more comments
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I])gives the same numerical result asPi/2Erf[Exp[-I Pi/4]]
– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
many thx for your solution
– Wolfdale
Jul 20 at 17:23
many thx for your solution
– Wolfdale
Jul 20 at 17:23
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
This answer gives $1.04460550032123928561$, which is not the same as what I got in two different ways. I also computed the original integral numerically and my answer checks out. The problem is probably minor since it appears the argument to erfc is negated.
– robjohn♦
Jul 21 at 14:02
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@robjohn I edited my answer. But I still can't figure out why Wolfy does not produce the correct result with the lower gamma function. Moreover, I don't know using what argument would make the lower gamma function to evaluate properly.
– Szeto
Jul 21 at 15:04
@Szeto: In Mathematica,
Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I]) gives the same numerical result as Pi/2Erf[Exp[-I Pi/4]]– robjohn♦
Jul 21 at 15:51
@Szeto: In Mathematica,
Sqrt[Pi]/2(Sqrt[Pi]-Gamma[1/2,-I]) gives the same numerical result as Pi/2Erf[Exp[-I Pi/4]]– robjohn♦
Jul 21 at 15:51
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
@robjohn I used the online version of WA. Do you think that will work also?
– Szeto
Jul 21 at 15:55
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Mathematica gives: $-frac12 pi left(sin (1) left(Sleft(sqrtfrac2pi right)-Cleft(sqrtfrac2pi right)right)+cos (1) left(Cleft(sqrtfrac2pi right)+Sleft(sqrtfrac2pi right)-1right)right)$
– David G. Stork
Jul 20 at 5:08
@DavidG.Stork, Yeah, I've got the same result by Wolfram Alpha. So, I tried to convert to Fresnel integral but in vain.
– Wolfdale
Jul 20 at 5:11
Fresnel integrals cannot be simplified, in general.
– David G. Stork
Jul 20 at 5:12
@DavidG.Stork but maybe a combination of them can be
– mathworker21
Jul 20 at 5:12
I strongly doubt it... otherwise I would likely have seen some identities to that effect. But I cannot prove this assertion without extensive effort.
– David G. Stork
Jul 20 at 5:14