Integral including functions $operatornameerfc(.), exp(.) $ and $ cos(.)$
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I have following integral. MATHEMATICA evaluates it as follows for $a>0$:
$$I=int_0^pi/2cos (theta ) e^a^2 cos ^2(theta ) texterfc(a cos (theta ))dtheta=fracsqrtpi left(1-e^a^2 texterfcleft(aright)right)2 a$$
However, I have no clue how this result comes. I have checked few integral table books such as Table of Integrals, Series, and Products, and also functions.wolfram.com. But I could not find matching expressions.
Does anyone have an idea?
integration analysis
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
asked Sep 2 at 6:24
Frey
626312
add a comment |Â
up vote
2
down vote
favorite
I have following integral. MATHEMATICA evaluates it as follows for $a>0$:
$$I=int_0^pi/2cos (theta ) e^a^2 cos ^2(theta ) texterfc(a cos (theta ))dtheta=fracsqrtpi left(1-e^a^2 texterfcleft(aright)right)2 a$$
However, I have no clue how this result comes. I have checked few integral table books such as Table of Integrals, Series, and Products, and also functions.wolfram.com. But I could not find matching expressions.
Does anyone have an idea?
integration analysis
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
asked Sep 2 at 6:24
Frey
626312
Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
I have tried this substitution before.
– Frey
Sep 2 at 7:12
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have following integral. MATHEMATICA evaluates it as follows for $a>0$:
$$I=int_0^pi/2cos (theta ) e^a^2 cos ^2(theta ) texterfc(a cos (theta ))dtheta=fracsqrtpi left(1-e^a^2 texterfcleft(aright)right)2 a$$
However, I have no clue how this result comes. I have checked few integral table books such as Table of Integrals, Series, and Products, and also functions.wolfram.com. But I could not find matching expressions.
Does anyone have an idea?
integration analysis
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
asked Sep 2 at 6:24
Frey
626312
I have following integral. MATHEMATICA evaluates it as follows for $a>0$:
$$I=int_0^pi/2cos (theta ) e^a^2 cos ^2(theta ) texterfc(a cos (theta ))dtheta=fracsqrtpi left(1-e^a^2 texterfcleft(aright)right)2 a$$
However, I have no clue how this result comes. I have checked few integral table books such as Table of Integrals, Series, and Products, and also functions.wolfram.com. But I could not find matching expressions.
Does anyone have an idea?
integration analysis
integration analysis
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
asked Sep 2 at 6:24
Frey
626312
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
asked Sep 2 at 6:24
Frey
626312
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
edited Sep 2 at 8:19
José Carlos Santos
121k16101186
121k16101186
asked Sep 2 at 6:24
Frey
626312
asked Sep 2 at 6:24
Frey
626312
asked Sep 2 at 6:24
Frey
626312
626312
Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
I have tried this substitution before.
– Frey
Sep 2 at 7:12
add a comment |Â
Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
I have tried this substitution before.
– Frey
Sep 2 at 7:12
Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
I have tried this substitution before.
– Frey
Sep 2 at 7:12
I have tried this substitution before.
– Frey
Sep 2 at 7:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
beginequation
int_0^inftyfrace^-a^2tsqrtt+cos^2thetamathrmdt=fracsqrtpiae^a^2cos^2thetaoperatornameerfcleft(acosthetaright) tag1label1
endequation
the integral can be transformed into
beginequation
I=fracasqrtpiint_0^pi/2cos theta ,dtheta int_0^inftyfrace^-a^2tsqrtt+cos^2theta,mathrmdt
endequation
By changing the integration order,
beginalign
I&=fracasqrtpiint_0^inftye^-a^2t,dtint_0^pi/2fraccos theta sqrtt+1-sin^2theta,dtheta\
&=fracasqrtpiint_0^inftye^-a^2tarcsin frac1sqrtt+1,dt
endalign
Now, integrating by parts, ($u'=e^-a^2t,v=arcsin frac1sqrtt+1$), we get
beginalign
I&=fracasqrtpileft[fracpi2a^2-frac12a^2int_0^inftyfrace^-a^2tsqrtt(t+1) right]\
&=fracasqrtpileft[fracpi2a^2-frac1a^2int_0^inftyfrace^-a^2u^2u^2+1 right]
endalign
(by changing $t=u^2$). Using the integral representation DLMF
beginequation
operatornameerfc(a)=frac2pie^-a^2int_0^inftyfrace^-a^2u^%
2u^2+1mathrmdu tag2label2
endequation
the final result is obtained:
beginequation
I= fracsqrtpi left(1-e^a^2 operatornameerfcleft(aright)right)2 a
endequation
eqref1 This identity is retrieved by changing $t=u^2-cos^2theta$ in the integral.
eqref2
With $A=a^2$, defining
beginalign
J(A)&=frac2piint_0^inftyfrace^-Aleft( u^2+1 right)u^2+1\
fracdJ(A)dA&=-frac2piint_0^infty e^-Aleft( u^2+1 right),du\
&=-frac1sqrtpi Ae^-A
endalign
and remarking that $J(0)=1$,
beginalign
J(A)&=1-frac1sqrtpiint_0^Afrace^-ssqrts,ds\
&=1-operatornameerf(a)
endalign
answered Sep 2 at 17:05
Paul Enta
3,3701925
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
beginequation
int_0^inftyfrace^-a^2tsqrtt+cos^2thetamathrmdt=fracsqrtpiae^a^2cos^2thetaoperatornameerfcleft(acosthetaright) tag1label1
endequation
the integral can be transformed into
beginequation
I=fracasqrtpiint_0^pi/2cos theta ,dtheta int_0^inftyfrace^-a^2tsqrtt+cos^2theta,mathrmdt
endequation
By changing the integration order,
beginalign
I&=fracasqrtpiint_0^inftye^-a^2t,dtint_0^pi/2fraccos theta sqrtt+1-sin^2theta,dtheta\
&=fracasqrtpiint_0^inftye^-a^2tarcsin frac1sqrtt+1,dt
endalign
Now, integrating by parts, ($u'=e^-a^2t,v=arcsin frac1sqrtt+1$), we get
beginalign
I&=fracasqrtpileft[fracpi2a^2-frac12a^2int_0^inftyfrace^-a^2tsqrtt(t+1) right]\
&=fracasqrtpileft[fracpi2a^2-frac1a^2int_0^inftyfrace^-a^2u^2u^2+1 right]
endalign
(by changing $t=u^2$). Using the integral representation DLMF
beginequation
operatornameerfc(a)=frac2pie^-a^2int_0^inftyfrace^-a^2u^%
2u^2+1mathrmdu tag2label2
endequation
the final result is obtained:
beginequation
I= fracsqrtpi left(1-e^a^2 operatornameerfcleft(aright)right)2 a
endequation
eqref1 This identity is retrieved by changing $t=u^2-cos^2theta$ in the integral.
eqref2
With $A=a^2$, defining
beginalign
J(A)&=frac2piint_0^inftyfrace^-Aleft( u^2+1 right)u^2+1\
fracdJ(A)dA&=-frac2piint_0^infty e^-Aleft( u^2+1 right),du\
&=-frac1sqrtpi Ae^-A
endalign
and remarking that $J(0)=1$,
beginalign
J(A)&=1-frac1sqrtpiint_0^Afrace^-ssqrts,ds\
&=1-operatornameerf(a)
endalign
answered Sep 2 at 17:05
Paul Enta
3,3701925
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
add a comment |Â
up vote
0
down vote
accepted
Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
beginequation
int_0^inftyfrace^-a^2tsqrtt+cos^2thetamathrmdt=fracsqrtpiae^a^2cos^2thetaoperatornameerfcleft(acosthetaright) tag1label1
endequation
the integral can be transformed into
beginequation
I=fracasqrtpiint_0^pi/2cos theta ,dtheta int_0^inftyfrace^-a^2tsqrtt+cos^2theta,mathrmdt
endequation
By changing the integration order,
beginalign
I&=fracasqrtpiint_0^inftye^-a^2t,dtint_0^pi/2fraccos theta sqrtt+1-sin^2theta,dtheta\
&=fracasqrtpiint_0^inftye^-a^2tarcsin frac1sqrtt+1,dt
endalign
Now, integrating by parts, ($u'=e^-a^2t,v=arcsin frac1sqrtt+1$), we get
beginalign
I&=fracasqrtpileft[fracpi2a^2-frac12a^2int_0^inftyfrace^-a^2tsqrtt(t+1) right]\
&=fracasqrtpileft[fracpi2a^2-frac1a^2int_0^inftyfrace^-a^2u^2u^2+1 right]
endalign
(by changing $t=u^2$). Using the integral representation DLMF
beginequation
operatornameerfc(a)=frac2pie^-a^2int_0^inftyfrace^-a^2u^%
2u^2+1mathrmdu tag2label2
endequation
the final result is obtained:
beginequation
I= fracsqrtpi left(1-e^a^2 operatornameerfcleft(aright)right)2 a
endequation
eqref1 This identity is retrieved by changing $t=u^2-cos^2theta$ in the integral.
eqref2
With $A=a^2$, defining
beginalign
J(A)&=frac2piint_0^inftyfrace^-Aleft( u^2+1 right)u^2+1\
fracdJ(A)dA&=-frac2piint_0^infty e^-Aleft( u^2+1 right),du\
&=-frac1sqrtpi Ae^-A
endalign
and remarking that $J(0)=1$,
beginalign
J(A)&=1-frac1sqrtpiint_0^Afrace^-ssqrts,ds\
&=1-operatornameerf(a)
endalign
answered Sep 2 at 17:05
Paul Enta
3,3701925
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
beginequation
int_0^inftyfrace^-a^2tsqrtt+cos^2thetamathrmdt=fracsqrtpiae^a^2cos^2thetaoperatornameerfcleft(acosthetaright) tag1label1
endequation
the integral can be transformed into
beginequation
I=fracasqrtpiint_0^pi/2cos theta ,dtheta int_0^inftyfrace^-a^2tsqrtt+cos^2theta,mathrmdt
endequation
By changing the integration order,
beginalign
I&=fracasqrtpiint_0^inftye^-a^2t,dtint_0^pi/2fraccos theta sqrtt+1-sin^2theta,dtheta\
&=fracasqrtpiint_0^inftye^-a^2tarcsin frac1sqrtt+1,dt
endalign
Now, integrating by parts, ($u'=e^-a^2t,v=arcsin frac1sqrtt+1$), we get
beginalign
I&=fracasqrtpileft[fracpi2a^2-frac12a^2int_0^inftyfrace^-a^2tsqrtt(t+1) right]\
&=fracasqrtpileft[fracpi2a^2-frac1a^2int_0^inftyfrace^-a^2u^2u^2+1 right]
endalign
(by changing $t=u^2$). Using the integral representation DLMF
beginequation
operatornameerfc(a)=frac2pie^-a^2int_0^inftyfrace^-a^2u^%
2u^2+1mathrmdu tag2label2
endequation
the final result is obtained:
beginequation
I= fracsqrtpi left(1-e^a^2 operatornameerfcleft(aright)right)2 a
endequation
eqref1 This identity is retrieved by changing $t=u^2-cos^2theta$ in the integral.
eqref2
With $A=a^2$, defining
beginalign
J(A)&=frac2piint_0^inftyfrace^-Aleft( u^2+1 right)u^2+1\
fracdJ(A)dA&=-frac2piint_0^infty e^-Aleft( u^2+1 right),du\
&=-frac1sqrtpi Ae^-A
endalign
and remarking that $J(0)=1$,
beginalign
J(A)&=1-frac1sqrtpiint_0^Afrace^-ssqrts,ds\
&=1-operatornameerf(a)
endalign
answered Sep 2 at 17:05
Paul Enta
3,3701925
Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
beginequation
int_0^inftyfrace^-a^2tsqrtt+cos^2thetamathrmdt=fracsqrtpiae^a^2cos^2thetaoperatornameerfcleft(acosthetaright) tag1label1
endequation
the integral can be transformed into
beginequation
I=fracasqrtpiint_0^pi/2cos theta ,dtheta int_0^inftyfrace^-a^2tsqrtt+cos^2theta,mathrmdt
endequation
By changing the integration order,
beginalign
I&=fracasqrtpiint_0^inftye^-a^2t,dtint_0^pi/2fraccos theta sqrtt+1-sin^2theta,dtheta\
&=fracasqrtpiint_0^inftye^-a^2tarcsin frac1sqrtt+1,dt
endalign
Now, integrating by parts, ($u'=e^-a^2t,v=arcsin frac1sqrtt+1$), we get
beginalign
I&=fracasqrtpileft[fracpi2a^2-frac12a^2int_0^inftyfrace^-a^2tsqrtt(t+1) right]\
&=fracasqrtpileft[fracpi2a^2-frac1a^2int_0^inftyfrace^-a^2u^2u^2+1 right]
endalign
(by changing $t=u^2$). Using the integral representation DLMF
beginequation
operatornameerfc(a)=frac2pie^-a^2int_0^inftyfrace^-a^2u^%
2u^2+1mathrmdu tag2label2
endequation
the final result is obtained:
beginequation
I= fracsqrtpi left(1-e^a^2 operatornameerfcleft(aright)right)2 a
endequation
eqref1 This identity is retrieved by changing $t=u^2-cos^2theta$ in the integral.
eqref2
With $A=a^2$, defining
beginalign
J(A)&=frac2piint_0^inftyfrace^-Aleft( u^2+1 right)u^2+1\
fracdJ(A)dA&=-frac2piint_0^infty e^-Aleft( u^2+1 right),du\
&=-frac1sqrtpi Ae^-A
endalign
and remarking that $J(0)=1$,
beginalign
J(A)&=1-frac1sqrtpiint_0^Afrace^-ssqrts,ds\
&=1-operatornameerf(a)
endalign
answered Sep 2 at 17:05
Paul Enta
3,3701925
edited Sep 5 at 19:26
answered Sep 2 at 17:05
Paul Enta
3,3701925
answered Sep 2 at 17:05
Paul Enta
3,3701925
answered Sep 2 at 17:05
Paul Enta
3,3701925
3,3701925
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
add a comment |Â
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
Really helpful, great! thanks @Paul.
– Frey
Sep 4 at 23:51
add a comment |Â
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Of course the most natural substitution is $x=acos(theta)$, which simplifies the integral by a lot. Otherwise you could use the classical en.wikipedia.org/wiki/Tangent_half-angle_substitution to avoid having roots in your integral. In any case, after that you probably need some partial integration to simply further.
– b00n heT
Sep 2 at 7:07
I have tried this substitution before.
– Frey
Sep 2 at 7:12