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Do these integrals evaluate to terms involving Kronecker Deltas?
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There is a famous integral representation of the Kronecker delta:
$$
delta_N,M = int_0^1 dx e^-2 pi i (N-M)x
$$
Noting this, I have encountered two integrals where $N,M,m in mathbbZ$. First:
$$
mathcalI_1 = int_m-tfrac12^m+tfrac12 dx x e^-2 pi i (N-M)x
$$
I have a reference which seems to be implying that $mathcalI_1 = m delta_M,N$.
The second integral is worse (for any $y > 0$):
$$
mathcalI_2 = int_m-tfrac12^m+tfrac12 dx sqrt x^2 + y^2 e^-2 pi i (N-M)x
$$
I have no clue how to evaluate either of these. Do they both involve a $delta_N,M$?
integration kronecker-delta
asked Aug 5 at 23:13
Greg.Paul
717620
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up vote
2
down vote
favorite
There is a famous integral representation of the Kronecker delta:
$$
delta_N,M = int_0^1 dx e^-2 pi i (N-M)x
$$
Noting this, I have encountered two integrals where $N,M,m in mathbbZ$. First:
$$
mathcalI_1 = int_m-tfrac12^m+tfrac12 dx x e^-2 pi i (N-M)x
$$
I have a reference which seems to be implying that $mathcalI_1 = m delta_M,N$.
The second integral is worse (for any $y > 0$):
$$
mathcalI_2 = int_m-tfrac12^m+tfrac12 dx sqrt x^2 + y^2 e^-2 pi i (N-M)x
$$
I have no clue how to evaluate either of these. Do they both involve a $delta_N,M$?
integration kronecker-delta
asked Aug 5 at 23:13
Greg.Paul
717620
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
There is a famous integral representation of the Kronecker delta:
$$
delta_N,M = int_0^1 dx e^-2 pi i (N-M)x
$$
Noting this, I have encountered two integrals where $N,M,m in mathbbZ$. First:
$$
mathcalI_1 = int_m-tfrac12^m+tfrac12 dx x e^-2 pi i (N-M)x
$$
I have a reference which seems to be implying that $mathcalI_1 = m delta_M,N$.
The second integral is worse (for any $y > 0$):
$$
mathcalI_2 = int_m-tfrac12^m+tfrac12 dx sqrt x^2 + y^2 e^-2 pi i (N-M)x
$$
I have no clue how to evaluate either of these. Do they both involve a $delta_N,M$?
integration kronecker-delta
asked Aug 5 at 23:13
Greg.Paul
717620
There is a famous integral representation of the Kronecker delta:
$$
delta_N,M = int_0^1 dx e^-2 pi i (N-M)x
$$
Noting this, I have encountered two integrals where $N,M,m in mathbbZ$. First:
$$
mathcalI_1 = int_m-tfrac12^m+tfrac12 dx x e^-2 pi i (N-M)x
$$
I have a reference which seems to be implying that $mathcalI_1 = m delta_M,N$.
The second integral is worse (for any $y > 0$):
$$
mathcalI_2 = int_m-tfrac12^m+tfrac12 dx sqrt x^2 + y^2 e^-2 pi i (N-M)x
$$
I have no clue how to evaluate either of these. Do they both involve a $delta_N,M$?
integration kronecker-delta
asked Aug 5 at 23:13
Greg.Paul
717620
asked Aug 5 at 23:13
Greg.Paul
717620
asked Aug 5 at 23:13
Greg.Paul
717620
asked Aug 5 at 23:13
Greg.Paul
717620
717620
add a comment |Â
add a comment |Â
1 Answer
1
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First note that your representation of the Kronecker delta is still correct if we shift the limits of integration:
$$ delta_N,M = int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x , .$$
For the first integral we get
$$ mathcalI_1 = int limits_m-frac12^m+frac12 mathrmd x , x = m $$
if $N=M$ . If $N neq M$, we can integrate by parts to obtain
beginalign
mathcalI_1 &= left[- fracx mathrme^- 2 pi mathrmi (N-M) x2 pi mathrmi (N-M)right]_x=m-frac12^x=m+frac12 + frac12 pi mathrmi (N-M)int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x \
&= fracmathrmi (-1)^N-M2 pi (N-M) , ,
endalign
since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $operatornameRe (mathcalI_1) = m delta_N,M$ holds, however.
The second integral is a lot more complicated, but numerical calculations suggest that neither real nor imaginary part vanish in general, so it seems unlikely that it has any connection to the Kronecker delta.
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First note that your representation of the Kronecker delta is still correct if we shift the limits of integration:
$$ delta_N,M = int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x , .$$
For the first integral we get
$$ mathcalI_1 = int limits_m-frac12^m+frac12 mathrmd x , x = m $$
if $N=M$ . If $N neq M$, we can integrate by parts to obtain
beginalign
mathcalI_1 &= left[- fracx mathrme^- 2 pi mathrmi (N-M) x2 pi mathrmi (N-M)right]_x=m-frac12^x=m+frac12 + frac12 pi mathrmi (N-M)int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x \
&= fracmathrmi (-1)^N-M2 pi (N-M) , ,
endalign
since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $operatornameRe (mathcalI_1) = m delta_N,M$ holds, however.
The second integral is a lot more complicated, but numerical calculations suggest that neither real nor imaginary part vanish in general, so it seems unlikely that it has any connection to the Kronecker delta.
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
add a comment |Â
up vote
2
down vote
accepted
First note that your representation of the Kronecker delta is still correct if we shift the limits of integration:
$$ delta_N,M = int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x , .$$
For the first integral we get
$$ mathcalI_1 = int limits_m-frac12^m+frac12 mathrmd x , x = m $$
if $N=M$ . If $N neq M$, we can integrate by parts to obtain
beginalign
mathcalI_1 &= left[- fracx mathrme^- 2 pi mathrmi (N-M) x2 pi mathrmi (N-M)right]_x=m-frac12^x=m+frac12 + frac12 pi mathrmi (N-M)int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x \
&= fracmathrmi (-1)^N-M2 pi (N-M) , ,
endalign
since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $operatornameRe (mathcalI_1) = m delta_N,M$ holds, however.
The second integral is a lot more complicated, but numerical calculations suggest that neither real nor imaginary part vanish in general, so it seems unlikely that it has any connection to the Kronecker delta.
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First note that your representation of the Kronecker delta is still correct if we shift the limits of integration:
$$ delta_N,M = int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x , .$$
For the first integral we get
$$ mathcalI_1 = int limits_m-frac12^m+frac12 mathrmd x , x = m $$
if $N=M$ . If $N neq M$, we can integrate by parts to obtain
beginalign
mathcalI_1 &= left[- fracx mathrme^- 2 pi mathrmi (N-M) x2 pi mathrmi (N-M)right]_x=m-frac12^x=m+frac12 + frac12 pi mathrmi (N-M)int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x \
&= fracmathrmi (-1)^N-M2 pi (N-M) , ,
endalign
since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $operatornameRe (mathcalI_1) = m delta_N,M$ holds, however.
The second integral is a lot more complicated, but numerical calculations suggest that neither real nor imaginary part vanish in general, so it seems unlikely that it has any connection to the Kronecker delta.
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
First note that your representation of the Kronecker delta is still correct if we shift the limits of integration:
$$ delta_N,M = int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x , .$$
For the first integral we get
$$ mathcalI_1 = int limits_m-frac12^m+frac12 mathrmd x , x = m $$
if $N=M$ . If $N neq M$, we can integrate by parts to obtain
beginalign
mathcalI_1 &= left[- fracx mathrme^- 2 pi mathrmi (N-M) x2 pi mathrmi (N-M)right]_x=m-frac12^x=m+frac12 + frac12 pi mathrmi (N-M)int limits_m-frac12^m+frac12 mathrmd x , mathrme^- 2 pi mathrmi (N-M) x \
&= fracmathrmi (-1)^N-M2 pi (N-M) , ,
endalign
since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $operatornameRe (mathcalI_1) = m delta_N,M$ holds, however.
The second integral is a lot more complicated, but numerical calculations suggest that neither real nor imaginary part vanish in general, so it seems unlikely that it has any connection to the Kronecker delta.
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
answered Aug 6 at 15:40
ComplexYetTrivial
2,817624
2,817624
add a comment |Â
add a comment |Â
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