Integrating a real nasty!Nasty Integral - Closed form solution?Integral evaluation $int_-infty^inftyfraccos (ax)pi (1+x^2)dx$Complex contour integralsMethod for integrating $int_-infty^infty x e^frac-x^22 dx$?Can this integral be evaluated/approximated?Integrating a Linear Operator $A:Hlongrightarrow H$ (Matrix)Is this way of integrating correct?Are there real definite integrals which can only be evaluated by contour integration?Real integrals in the complex planeHow can I use residues to solve this real integral?
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Integrating a real nasty!
Nasty Integral - Closed form solution?Integral evaluation $int_-infty^inftyfraccos (ax)pi (1+x^2)dx$Complex contour integralsMethod for integrating $int_-infty^infty x e^frac-x^22 dx$?Can this integral be evaluated/approximated?Integrating a Linear Operator $A:Hlongrightarrow H$ (Matrix)Is this way of integrating correct?Are there real definite integrals which can only be evaluated by contour integration?Real integrals in the complex planeHow can I use residues to solve this real integral?
$begingroup$
I came across this beast
$$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$
where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?
integration
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
$endgroup$
add a comment |
$begingroup$
I came across this beast
$$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$
where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?
integration
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
$endgroup$
add a comment |
$begingroup$
I came across this beast
$$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$
where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?
integration
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
$endgroup$
I came across this beast
$$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$
where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?
integration
integration
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
asked Mar 28 at 15:57
keeran_q789keeran_q789
184
184
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $|a|<1$, we have
$$
I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
$$ Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
$$beginalign*
I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
endalign*$$ which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
$$
I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
$$ to obtain
$$
I(a)=-fracpi a^-nn.
$$
answered Mar 28 at 16:41
NaoNao
1836
$endgroup$
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
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
$begingroup$
For $|a|<1$, we have
$$
I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
$$ Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
$$beginalign*
I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
endalign*$$ which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
$$
I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
$$ to obtain
$$
I(a)=-fracpi a^-nn.
$$
answered Mar 28 at 16:41
NaoNao
1836
$endgroup$
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
add a comment |
$begingroup$
For $|a|<1$, we have
$$
I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
$$ Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
$$beginalign*
I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
endalign*$$ which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
$$
I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
$$ to obtain
$$
I(a)=-fracpi a^-nn.
$$
answered Mar 28 at 16:41
NaoNao
1836
$endgroup$
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
add a comment |
$begingroup$
For $|a|<1$, we have
$$
I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
$$ Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
$$beginalign*
I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
endalign*$$ which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
$$
I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
$$ to obtain
$$
I(a)=-fracpi a^-nn.
$$
answered Mar 28 at 16:41
NaoNao
1836
$endgroup$
For $|a|<1$, we have
$$
I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
$$ Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
$$beginalign*
I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
endalign*$$ which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
$$
I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
$$ to obtain
$$
I(a)=-fracpi a^-nn.
$$
answered Mar 28 at 16:41
NaoNao
1836
answered Mar 28 at 16:41
NaoNao
1836
answered Mar 28 at 16:41
NaoNao
1836
answered Mar 28 at 16:41
NaoNao
1836
1836
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
add a comment |
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
1
1
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
Nice (+1). Where can one see a proof of that interesting Fourier series?
$endgroup$
– clathratus
Mar 28 at 16:47
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
$begingroup$
@clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
$endgroup$
– Nao
Mar 28 at 16:51
add a comment |
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