Mixing
Calculate $intlimits_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx$ using complex analysis technique
$begingroup$
In a complex analysis test an exercise asks to calculate ($w in Bbb{R}$): $$displaystyleint_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx$$
Of course I need to solve it with complex analysis technique. I have used all the tricks I know such as integrate on a semi-circumference of radius $R$ in $Bbb{C}$ and then let $R to infty$ but everything I have tried to do just seemed useless. Thank you for every hint or solution to this problem!
integration complex-analysis definite-integrals improper-integrals
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
$endgroup$
add a comment |
$begingroup$
In a complex analysis test an exercise asks to calculate ($w in Bbb{R}$): $$displaystyleint_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx$$
Of course I need to solve it with complex analysis technique. I have used all the tricks I know such as integrate on a semi-circumference of radius $R$ in $Bbb{C}$ and then let $R to infty$ but everything I have tried to do just seemed useless. Thank you for every hint or solution to this problem!
integration complex-analysis definite-integrals improper-integrals
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
$endgroup$
1
$begingroup$
Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08
add a comment |
$begingroup$
In a complex analysis test an exercise asks to calculate ($w in Bbb{R}$): $$displaystyleint_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx$$
Of course I need to solve it with complex analysis technique. I have used all the tricks I know such as integrate on a semi-circumference of radius $R$ in $Bbb{C}$ and then let $R to infty$ but everything I have tried to do just seemed useless. Thank you for every hint or solution to this problem!
integration complex-analysis definite-integrals improper-integrals
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
$endgroup$
In a complex analysis test an exercise asks to calculate ($w in Bbb{R}$): $$displaystyleint_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx$$
Of course I need to solve it with complex analysis technique. I have used all the tricks I know such as integrate on a semi-circumference of radius $R$ in $Bbb{C}$ and then let $R to infty$ but everything I have tried to do just seemed useless. Thank you for every hint or solution to this problem!
integration complex-analysis definite-integrals improper-integrals
integration complex-analysis definite-integrals improper-integrals
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
edited Jan 20 at 9:05
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 19 at 20:53
edo1998edo1998
419113
asked Jan 19 at 20:53
edo1998edo1998
419113
asked Jan 19 at 20:53
edo1998edo1998
419113
419113
1
$begingroup$
Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08
add a comment |
1
$begingroup$
Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08
1
1
$begingroup$
Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If I had to use complex methods in this, the place I'd do it is in proving$$int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx=pi$$(e.g. by the residue theorem; see here for the gory details) so$$left|kright|pi=int_{mathbb{R}}frac{sin^{2}kx}{x^{2}}dx=int_{mathbb{R}}frac{1-cos2kx}{2x^{2}}dx.$$Hence$$int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx\=frac{pi}{4}left(left|w-2right|+left|w+2right|-2left|wright|right).$$As a sanity check, the case $w=0$ needs to give $pi$, and lo and behold it does.
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
$endgroup$
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
|
show 2 more comments
$begingroup$
This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that
$$
int_{-1}^1 e^{-ixi x}dx=frac{2sin xi}{xi}.
$$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $widehat{f}(xi)=frac{2sin xi}{xi}$ and therefore, $widehat{(f*f)}(xi)=f(xi)^2=4frac{2sin^2 xi}{xi^2}$.
The given integral is
$$begin{eqnarray}
int_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx&=&int_{-infty}^{infty}frac{sin^2(x)e^{iwx}}{x^2}dx\
&=&2pi mathcal{F}^{-1}left[frac{sin^2 x}{x^2}right](w)\
&=&frac{pi}{2}(f*f)(w).
end{eqnarray}$$ Finally, we have
$$begin{eqnarray}
(f*f)(w)&=&int f(w-y)f(y)dy\
&=&int_{|w-y|<1,|y|<1}dy\
&=&begin{cases}2-w,quad win[0,2)\2+w,quad win(-2,0]\0,quad|w|>2end{cases}\
&=&(2-|w|)1_{|w|<2}.
end{eqnarray}$$ This gives that the given integral is equal to
$$
frac{pi}{2}(2-|w|)1_{|w|<2}.
$$
answered Jan 19 at 22:13
SongSong
16.2k1739
$endgroup$
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
add a comment |
$begingroup$
$$I(w)=int_{-infty}^inftyfrac{sin^2(x)cos(wx)}{x^2}dx$$
$$I'(w)=-int_{-infty}^inftyfrac{sin^2(x)sin(wx)}{x}$$
$$I''(w)=-int_{-infty}^inftysin^2(x)cos(wx)dx$$
$$I''(w)=-Reint_{-infty}^inftysin^2(x)e^{iwx}dx$$
and then try integration by parts. Alternatively:
$$I(w)=Reint_{-infty}^inftyfrac{sin^2(x)e^{iwx}}{x^2}dx$$
$$I(w)=Reint_{-infty}^inftysin^2(x)sum_{n=0}^inftyfrac{(iw)^nx^{n-2}}{n!}dx$$
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I had to use complex methods in this, the place I'd do it is in proving$$int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx=pi$$(e.g. by the residue theorem; see here for the gory details) so$$left|kright|pi=int_{mathbb{R}}frac{sin^{2}kx}{x^{2}}dx=int_{mathbb{R}}frac{1-cos2kx}{2x^{2}}dx.$$Hence$$int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx\=frac{pi}{4}left(left|w-2right|+left|w+2right|-2left|wright|right).$$As a sanity check, the case $w=0$ needs to give $pi$, and lo and behold it does.
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
$endgroup$
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
|
show 2 more comments
$begingroup$
If I had to use complex methods in this, the place I'd do it is in proving$$int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx=pi$$(e.g. by the residue theorem; see here for the gory details) so$$left|kright|pi=int_{mathbb{R}}frac{sin^{2}kx}{x^{2}}dx=int_{mathbb{R}}frac{1-cos2kx}{2x^{2}}dx.$$Hence$$int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx\=frac{pi}{4}left(left|w-2right|+left|w+2right|-2left|wright|right).$$As a sanity check, the case $w=0$ needs to give $pi$, and lo and behold it does.
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
$endgroup$
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
|
show 2 more comments
$begingroup$
If I had to use complex methods in this, the place I'd do it is in proving$$int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx=pi$$(e.g. by the residue theorem; see here for the gory details) so$$left|kright|pi=int_{mathbb{R}}frac{sin^{2}kx}{x^{2}}dx=int_{mathbb{R}}frac{1-cos2kx}{2x^{2}}dx.$$Hence$$int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx\=frac{pi}{4}left(left|w-2right|+left|w+2right|-2left|wright|right).$$As a sanity check, the case $w=0$ needs to give $pi$, and lo and behold it does.
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
$endgroup$
If I had to use complex methods in this, the place I'd do it is in proving$$int_{mathbb{R}}frac{sin^{2}x}{x^{2}}dx=pi$$(e.g. by the residue theorem; see here for the gory details) so$$left|kright|pi=int_{mathbb{R}}frac{sin^{2}kx}{x^{2}}dx=int_{mathbb{R}}frac{1-cos2kx}{2x^{2}}dx.$$Hence$$int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx\=frac{pi}{4}left(left|w-2right|+left|w+2right|-2left|wright|right).$$As a sanity check, the case $w=0$ needs to give $pi$, and lo and behold it does.
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
answered Jan 19 at 21:54
J.G.J.G.
28.4k22845
28.4k22845
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
|
show 2 more comments
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
$begingroup$
What have you done in this equality? $int_{mathbb{R}}frac{sin^{2}xcos wx}{x^{2}}dx=int_{mathbb{R}}frac{2cos wx-cosleft(w-2right)x-cosleft(w+2right)x}{4x^{2}}dx$
$endgroup$
– edo1998
Jan 19 at 22:01
1
1
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
@edo1998 $sin^2x=(1-cos 2x)/2$ followed by the product-of-two-cosines-as-sum-of-two-cosines formula.
$endgroup$
– J.G.
Jan 19 at 22:02
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
Can you please explain in more detail how did the absolute values appear in the answer.
$endgroup$
– user
Jan 19 at 22:19
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
@user While the cases $k=0,,k>0$ are separately trivial, the obvious substitution $y=kx$ induces a sign change if $k<0$ due to exchanging the $pminfty$ limits. Another way to see it is the integrand is even in $k$.
$endgroup$
– J.G.
Jan 19 at 22:22
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
$begingroup$
I know that the absolute value is the correct result. From the attached drawing it is however not clear how to obtain it from the residue theorem. :)
$endgroup$
– user
Jan 19 at 22:33
|
show 2 more comments
$begingroup$
This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that
$$
int_{-1}^1 e^{-ixi x}dx=frac{2sin xi}{xi}.
$$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $widehat{f}(xi)=frac{2sin xi}{xi}$ and therefore, $widehat{(f*f)}(xi)=f(xi)^2=4frac{2sin^2 xi}{xi^2}$.
The given integral is
$$begin{eqnarray}
int_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx&=&int_{-infty}^{infty}frac{sin^2(x)e^{iwx}}{x^2}dx\
&=&2pi mathcal{F}^{-1}left[frac{sin^2 x}{x^2}right](w)\
&=&frac{pi}{2}(f*f)(w).
end{eqnarray}$$ Finally, we have
$$begin{eqnarray}
(f*f)(w)&=&int f(w-y)f(y)dy\
&=&int_{|w-y|<1,|y|<1}dy\
&=&begin{cases}2-w,quad win[0,2)\2+w,quad win(-2,0]\0,quad|w|>2end{cases}\
&=&(2-|w|)1_{|w|<2}.
end{eqnarray}$$ This gives that the given integral is equal to
$$
frac{pi}{2}(2-|w|)1_{|w|<2}.
$$
answered Jan 19 at 22:13
SongSong
16.2k1739
$endgroup$
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
add a comment |
$begingroup$
This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that
$$
int_{-1}^1 e^{-ixi x}dx=frac{2sin xi}{xi}.
$$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $widehat{f}(xi)=frac{2sin xi}{xi}$ and therefore, $widehat{(f*f)}(xi)=f(xi)^2=4frac{2sin^2 xi}{xi^2}$.
The given integral is
$$begin{eqnarray}
int_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx&=&int_{-infty}^{infty}frac{sin^2(x)e^{iwx}}{x^2}dx\
&=&2pi mathcal{F}^{-1}left[frac{sin^2 x}{x^2}right](w)\
&=&frac{pi}{2}(f*f)(w).
end{eqnarray}$$ Finally, we have
$$begin{eqnarray}
(f*f)(w)&=&int f(w-y)f(y)dy\
&=&int_{|w-y|<1,|y|<1}dy\
&=&begin{cases}2-w,quad win[0,2)\2+w,quad win(-2,0]\0,quad|w|>2end{cases}\
&=&(2-|w|)1_{|w|<2}.
end{eqnarray}$$ This gives that the given integral is equal to
$$
frac{pi}{2}(2-|w|)1_{|w|<2}.
$$
answered Jan 19 at 22:13
SongSong
16.2k1739
$endgroup$
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
add a comment |
$begingroup$
This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that
$$
int_{-1}^1 e^{-ixi x}dx=frac{2sin xi}{xi}.
$$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $widehat{f}(xi)=frac{2sin xi}{xi}$ and therefore, $widehat{(f*f)}(xi)=f(xi)^2=4frac{2sin^2 xi}{xi^2}$.
The given integral is
$$begin{eqnarray}
int_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx&=&int_{-infty}^{infty}frac{sin^2(x)e^{iwx}}{x^2}dx\
&=&2pi mathcal{F}^{-1}left[frac{sin^2 x}{x^2}right](w)\
&=&frac{pi}{2}(f*f)(w).
end{eqnarray}$$ Finally, we have
$$begin{eqnarray}
(f*f)(w)&=&int f(w-y)f(y)dy\
&=&int_{|w-y|<1,|y|<1}dy\
&=&begin{cases}2-w,quad win[0,2)\2+w,quad win(-2,0]\0,quad|w|>2end{cases}\
&=&(2-|w|)1_{|w|<2}.
end{eqnarray}$$ This gives that the given integral is equal to
$$
frac{pi}{2}(2-|w|)1_{|w|<2}.
$$
answered Jan 19 at 22:13
SongSong
16.2k1739
$endgroup$
This is certainly not a complex analytic method, but Fourier inversion formula also can give you the result. It holds that
$$
int_{-1}^1 e^{-ixi x}dx=frac{2sin xi}{xi}.
$$ Let $f(x) = 1_{[-1,1]}(x)$. This shows $widehat{f}(xi)=frac{2sin xi}{xi}$ and therefore, $widehat{(f*f)}(xi)=f(xi)^2=4frac{2sin^2 xi}{xi^2}$.
The given integral is
$$begin{eqnarray}
int_{-infty}^{infty}frac{sin^2(x)cos(wx)}{x^2}dx&=&int_{-infty}^{infty}frac{sin^2(x)e^{iwx}}{x^2}dx\
&=&2pi mathcal{F}^{-1}left[frac{sin^2 x}{x^2}right](w)\
&=&frac{pi}{2}(f*f)(w).
end{eqnarray}$$ Finally, we have
$$begin{eqnarray}
(f*f)(w)&=&int f(w-y)f(y)dy\
&=&int_{|w-y|<1,|y|<1}dy\
&=&begin{cases}2-w,quad win[0,2)\2+w,quad win(-2,0]\0,quad|w|>2end{cases}\
&=&(2-|w|)1_{|w|<2}.
end{eqnarray}$$ This gives that the given integral is equal to
$$
frac{pi}{2}(2-|w|)1_{|w|<2}.
$$
answered Jan 19 at 22:13
SongSong
16.2k1739
edited Jan 19 at 22:20
answered Jan 19 at 22:13
SongSong
16.2k1739
answered Jan 19 at 22:13
SongSong
16.2k1739
answered Jan 19 at 22:13
SongSong
16.2k1739
16.2k1739
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I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
add a comment |
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
$begingroup$
I'll look at your answer when I'll have the knowledge to understand it!
$endgroup$
– edo1998
Jan 19 at 22:31
1
1
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
$begingroup$
@edo1998 I hope it will help you soon! :)
$endgroup$
– Song
Jan 19 at 22:32
add a comment |
$begingroup$
$$I(w)=int_{-infty}^inftyfrac{sin^2(x)cos(wx)}{x^2}dx$$
$$I'(w)=-int_{-infty}^inftyfrac{sin^2(x)sin(wx)}{x}$$
$$I''(w)=-int_{-infty}^inftysin^2(x)cos(wx)dx$$
$$I''(w)=-Reint_{-infty}^inftysin^2(x)e^{iwx}dx$$
and then try integration by parts. Alternatively:
$$I(w)=Reint_{-infty}^inftyfrac{sin^2(x)e^{iwx}}{x^2}dx$$
$$I(w)=Reint_{-infty}^inftysin^2(x)sum_{n=0}^inftyfrac{(iw)^nx^{n-2}}{n!}dx$$
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
$$I(w)=int_{-infty}^inftyfrac{sin^2(x)cos(wx)}{x^2}dx$$
$$I'(w)=-int_{-infty}^inftyfrac{sin^2(x)sin(wx)}{x}$$
$$I''(w)=-int_{-infty}^inftysin^2(x)cos(wx)dx$$
$$I''(w)=-Reint_{-infty}^inftysin^2(x)e^{iwx}dx$$
and then try integration by parts. Alternatively:
$$I(w)=Reint_{-infty}^inftyfrac{sin^2(x)e^{iwx}}{x^2}dx$$
$$I(w)=Reint_{-infty}^inftysin^2(x)sum_{n=0}^inftyfrac{(iw)^nx^{n-2}}{n!}dx$$
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
$endgroup$
add a comment |
$begingroup$
$$I(w)=int_{-infty}^inftyfrac{sin^2(x)cos(wx)}{x^2}dx$$
$$I'(w)=-int_{-infty}^inftyfrac{sin^2(x)sin(wx)}{x}$$
$$I''(w)=-int_{-infty}^inftysin^2(x)cos(wx)dx$$
$$I''(w)=-Reint_{-infty}^inftysin^2(x)e^{iwx}dx$$
and then try integration by parts. Alternatively:
$$I(w)=Reint_{-infty}^inftyfrac{sin^2(x)e^{iwx}}{x^2}dx$$
$$I(w)=Reint_{-infty}^inftysin^2(x)sum_{n=0}^inftyfrac{(iw)^nx^{n-2}}{n!}dx$$
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
$endgroup$
$$I(w)=int_{-infty}^inftyfrac{sin^2(x)cos(wx)}{x^2}dx$$
$$I'(w)=-int_{-infty}^inftyfrac{sin^2(x)sin(wx)}{x}$$
$$I''(w)=-int_{-infty}^inftysin^2(x)cos(wx)dx$$
$$I''(w)=-Reint_{-infty}^inftysin^2(x)e^{iwx}dx$$
and then try integration by parts. Alternatively:
$$I(w)=Reint_{-infty}^inftyfrac{sin^2(x)e^{iwx}}{x^2}dx$$
$$I(w)=Reint_{-infty}^inftysin^2(x)sum_{n=0}^inftyfrac{(iw)^nx^{n-2}}{n!}dx$$
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
answered Jan 19 at 21:33
Henry LeeHenry Lee
2,054219
2,054219
add a comment |
add a comment |
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Hint: this function is holomorfic ( in particular it is defined in x=0 ).
$endgroup$
– Jakub Andruszkiewicz
Jan 19 at 21:05
$begingroup$
@JakubAndruszkiewicz I know, in $x=0$ there is a removable singularity but still I can't proceed, do you have any other hint?
$endgroup$
– edo1998
Jan 19 at 21:08