Using the Cauchy Integral Theorem for Derivatives to evaluate an integral
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I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?
$$f^{(k)}(w)=frac{k!}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
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add a comment |
$begingroup$
I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?
$$f^{(k)}(w)=frac{k!}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
$endgroup$
add a comment |
$begingroup$
I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?
$$f^{(k)}(w)=frac{k!}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
$endgroup$
I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz$$ I tried using the fact that $displaystyle int_gamma f(z)dz=int_a^b f(gamma(t))gamma'(t)dt$ for $gamma(t)$ where $t in [a, b]$. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?
$$f^{(k)}(w)=frac{k!}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
edited Jan 11 at 12:43
Paras Khosla
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
asked Nov 22 '18 at 16:49
Paras KhoslaParas Khosla
8610
8610
add a comment |
add a comment |
2 Answers
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By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
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The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
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2 Answers
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2 Answers
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$begingroup$
By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
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add a comment |
$begingroup$
By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
By Cauchy's integral theorem,$$oint_{lvert zrvert=2}frac{cos z}{z(z^2+8)},mathrm dz=oint_{lvert zrvert=2}frac{frac{cos z}{z^2+8}}z,mathrm dz=2pi itimesfrac{cos0}{0^2+8}=frac{pi i}4.$$
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
answered Nov 22 '18 at 16:52
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
add a comment |
add a comment |
$begingroup$
The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
$endgroup$
add a comment |
$begingroup$
The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
$endgroup$
The function $displaystyle f(z)=frac{cos(z)}{z^2+8}$ is analytic on $|z|=2$ then using Cauchy's Integral Theorem
$$displaystyle int_{|z|=2} frac{cos(z)}{z(z^2+8)}dz=int_{|z|=2} dfrac{frac{cos(z)}{z^2+8}}{z}dz=2pi itimes f(0)=dfrac{pi i}{4}$$
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
answered Nov 22 '18 at 16:53
NosratiNosrati
26.5k62354
26.5k62354
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