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How to evaluate this gaussian integral
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Is there a way to solve this integral?
$$int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds$$
I usually apply this formula for integrals of this kind
$$int_-infty^infty e^-(as^2+bs+c),ds=sqrtfracpiae^fracb^24a-c$$
Thank you in advance
calculus integration
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
asked 2 hours ago
Riccardo
263
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Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
4
down vote
favorite
Is there a way to solve this integral?
$$int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds$$
I usually apply this formula for integrals of this kind
$$int_-infty^infty e^-(as^2+bs+c),ds=sqrtfracpiae^fracb^24a-c$$
Thank you in advance
calculus integration
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
asked 2 hours ago
Riccardo
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
3
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Is there a way to solve this integral?
$$int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds$$
I usually apply this formula for integrals of this kind
$$int_-infty^infty e^-(as^2+bs+c),ds=sqrtfracpiae^fracb^24a-c$$
Thank you in advance
calculus integration
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
asked 2 hours ago
Riccardo
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there a way to solve this integral?
$$int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds$$
I usually apply this formula for integrals of this kind
$$int_-infty^infty e^-(as^2+bs+c),ds=sqrtfracpiae^fracb^24a-c$$
Thank you in advance
calculus integration
calculus integration
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
asked 2 hours ago
Riccardo
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
asked 2 hours ago
Riccardo
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
edited 1 hour ago
DarÃo A. Gutiérrez
2,52331430
2,52331430
asked 2 hours ago
Riccardo
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Riccardo
263
asked 2 hours ago
Riccardo
263
263
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Riccardo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
3
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago
add a comment |Â
3
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
3
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago
3
3
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
3
3
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,cinmathbb C$ provided that $mathrmRe[a] > 0$. It gives
beginmultline
int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds = mathrmImleft[int_-infty^infty e^-(s^2+2sqrtts)e^ileft(y+2sqrttsright)dsright]
\= mathrmImleft[int_-infty^infty e^-[s^2+2(1-i)sqrtts-iy]dsright] = mathrmImleft[sqrtpiexpleft(-2it +i yright)right] = sqrtpisin(y-2t).
endmultline
If you're not OK with working with complex numbers, there actually are formulas for these:
begineqnarray
int_-infty^infty e^-(as^2 + bs + c)sin(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - csinleft(m-fracb k2aright)\
int_-infty^infty e^-(as^2 + bs + c)cos(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - ccosleft(m-fracb k2aright)
endeqnarray
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $sqrtpisin(y-2t)$ again.
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
answered 59 mins ago
eyeballfrog
5,301528
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,cinmathbb C$ provided that $mathrmRe[a] > 0$. It gives
beginmultline
int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds = mathrmImleft[int_-infty^infty e^-(s^2+2sqrtts)e^ileft(y+2sqrttsright)dsright]
\= mathrmImleft[int_-infty^infty e^-[s^2+2(1-i)sqrtts-iy]dsright] = mathrmImleft[sqrtpiexpleft(-2it +i yright)right] = sqrtpisin(y-2t).
endmultline
If you're not OK with working with complex numbers, there actually are formulas for these:
begineqnarray
int_-infty^infty e^-(as^2 + bs + c)sin(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - csinleft(m-fracb k2aright)\
int_-infty^infty e^-(as^2 + bs + c)cos(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - ccosleft(m-fracb k2aright)
endeqnarray
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $sqrtpisin(y-2t)$ again.
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
answered 59 mins ago
eyeballfrog
5,301528
add a comment |Â
up vote
4
down vote
accepted
If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,cinmathbb C$ provided that $mathrmRe[a] > 0$. It gives
beginmultline
int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds = mathrmImleft[int_-infty^infty e^-(s^2+2sqrtts)e^ileft(y+2sqrttsright)dsright]
\= mathrmImleft[int_-infty^infty e^-[s^2+2(1-i)sqrtts-iy]dsright] = mathrmImleft[sqrtpiexpleft(-2it +i yright)right] = sqrtpisin(y-2t).
endmultline
If you're not OK with working with complex numbers, there actually are formulas for these:
begineqnarray
int_-infty^infty e^-(as^2 + bs + c)sin(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - csinleft(m-fracb k2aright)\
int_-infty^infty e^-(as^2 + bs + c)cos(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - ccosleft(m-fracb k2aright)
endeqnarray
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $sqrtpisin(y-2t)$ again.
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
answered 59 mins ago
eyeballfrog
5,301528
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,cinmathbb C$ provided that $mathrmRe[a] > 0$. It gives
beginmultline
int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds = mathrmImleft[int_-infty^infty e^-(s^2+2sqrtts)e^ileft(y+2sqrttsright)dsright]
\= mathrmImleft[int_-infty^infty e^-[s^2+2(1-i)sqrtts-iy]dsright] = mathrmImleft[sqrtpiexpleft(-2it +i yright)right] = sqrtpisin(y-2t).
endmultline
If you're not OK with working with complex numbers, there actually are formulas for these:
begineqnarray
int_-infty^infty e^-(as^2 + bs + c)sin(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - csinleft(m-fracb k2aright)\
int_-infty^infty e^-(as^2 + bs + c)cos(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - ccosleft(m-fracb k2aright)
endeqnarray
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $sqrtpisin(y-2t)$ again.
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
answered 59 mins ago
eyeballfrog
5,301528
If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,cinmathbb C$ provided that $mathrmRe[a] > 0$. It gives
beginmultline
int_-infty^infty e^-(s^2+2sqrtts)sin(y+2sqrtts),ds = mathrmImleft[int_-infty^infty e^-(s^2+2sqrtts)e^ileft(y+2sqrttsright)dsright]
\= mathrmImleft[int_-infty^infty e^-[s^2+2(1-i)sqrtts-iy]dsright] = mathrmImleft[sqrtpiexpleft(-2it +i yright)right] = sqrtpisin(y-2t).
endmultline
If you're not OK with working with complex numbers, there actually are formulas for these:
begineqnarray
int_-infty^infty e^-(as^2 + bs + c)sin(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - csinleft(m-fracb k2aright)\
int_-infty^infty e^-(as^2 + bs + c)cos(ks + m)ds &=& sqrtfracpiae^fracb^2-k^24a - ccosleft(m-fracb k2aright)
endeqnarray
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $sqrtpisin(y-2t)$ again.
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
answered 59 mins ago
eyeballfrog
5,301528
answered 59 mins ago
eyeballfrog
5,301528
answered 59 mins ago
eyeballfrog
5,301528
answered 59 mins ago
eyeballfrog
5,301528
5,301528
add a comment |Â
add a comment |Â
Riccardo is a new contributor. Be nice, and check out our Code of Conduct.
Riccardo is a new contributor. Be nice, and check out our Code of Conduct.
Riccardo is a new contributor. Be nice, and check out our Code of Conduct.
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3
Use $sin(y+2sqrtts)=frace^i(y+2sqrtts)-e^-i(y+2sqrtts)2i$ and then use your formula.
– aleden
2 hours ago
3
With complex integration, this is easy (your last formula is valid for any complex $b$ - as both sides represent entire functions of $b$ - and that's enough to get the answer right away).
– metamorphy
2 hours ago