Is $int_{0}^{infty} frac{sin^2 x }{x^2}dx$ equal to $int_{0}^{infty} frac{sin x }{x}dx$?
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I saw this in a solution to a problem and can't see why this would be the case. Tried substition and other techniques.
integration improper-integrals
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
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up vote
1
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I saw this in a solution to a problem and can't see why this would be the case. Tried substition and other techniques.
integration improper-integrals
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
I saw this in a solution to a problem and can't see why this would be the case. Tried substition and other techniques.
integration improper-integrals
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
I saw this in a solution to a problem and can't see why this would be the case. Tried substition and other techniques.
integration improper-integrals
integration improper-integrals
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
edited Nov 27 at 9:04
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
asked Nov 27 at 8:36
Nuntractatuses Amável
57212
57212
It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57
add a comment |
It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57
It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57
It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57
add a comment |
1 Answer
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accepted
$$
int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx = left[sin^{2}xleft(-frac{1}{x}right)right]^{infty}_{-infty} - int_{mathbb{R}} 2sin xcos x left(-frac{1}{x}right)dx \
=int_{mathbb{R}}frac{sin 2x}{x}dx = int_{mathbb{R}}frac{sin y}{y}dy
$$
Here we use integration by parts and the substitution $y = 2x$.
answered Nov 27 at 9:03
Seewoo Lee
6,110826
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
$$
int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx = left[sin^{2}xleft(-frac{1}{x}right)right]^{infty}_{-infty} - int_{mathbb{R}} 2sin xcos x left(-frac{1}{x}right)dx \
=int_{mathbb{R}}frac{sin 2x}{x}dx = int_{mathbb{R}}frac{sin y}{y}dy
$$
Here we use integration by parts and the substitution $y = 2x$.
answered Nov 27 at 9:03
Seewoo Lee
6,110826
add a comment |
up vote
7
down vote
accepted
$$
int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx = left[sin^{2}xleft(-frac{1}{x}right)right]^{infty}_{-infty} - int_{mathbb{R}} 2sin xcos x left(-frac{1}{x}right)dx \
=int_{mathbb{R}}frac{sin 2x}{x}dx = int_{mathbb{R}}frac{sin y}{y}dy
$$
Here we use integration by parts and the substitution $y = 2x$.
answered Nov 27 at 9:03
Seewoo Lee
6,110826
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
$$
int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx = left[sin^{2}xleft(-frac{1}{x}right)right]^{infty}_{-infty} - int_{mathbb{R}} 2sin xcos x left(-frac{1}{x}right)dx \
=int_{mathbb{R}}frac{sin 2x}{x}dx = int_{mathbb{R}}frac{sin y}{y}dy
$$
Here we use integration by parts and the substitution $y = 2x$.
answered Nov 27 at 9:03
Seewoo Lee
6,110826
$$
int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx = left[sin^{2}xleft(-frac{1}{x}right)right]^{infty}_{-infty} - int_{mathbb{R}} 2sin xcos x left(-frac{1}{x}right)dx \
=int_{mathbb{R}}frac{sin 2x}{x}dx = int_{mathbb{R}}frac{sin y}{y}dy
$$
Here we use integration by parts and the substitution $y = 2x$.
answered Nov 27 at 9:03
Seewoo Lee
6,110826
answered Nov 27 at 9:03
Seewoo Lee
6,110826
answered Nov 27 at 9:03
Seewoo Lee
6,110826
answered Nov 27 at 9:03
Seewoo Lee
6,110826
6,110826
add a comment |
add a comment |
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It can be a simple coincidence. Does this appear as a step in a proof ? Provide more context.
– Yves Daoust
Nov 27 at 8:57