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?
$$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 at 16:49
Paras Khosla
449
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up vote
0
down vote
favorite
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?
$$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 at 16:49
Paras Khosla
449
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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?
$$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
asked Nov 22 at 16:49
Paras Khosla
449
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?
$$f^{(k)}(w)=frac{1}{2pi i}int_gamma frac{f(z)}{(z-w)^{k+1}}dz$$ where $f^{(k)}(w)$ is the $k^{th}$ derivative of $f$
complex-analysis complex-integration
complex-analysis complex-integration
asked Nov 22 at 16:49
Paras Khosla
449
asked Nov 22 at 16:49
Paras Khosla
449
asked Nov 22 at 16:49
Paras Khosla
449
asked Nov 22 at 16:49
Paras Khosla
449
asked Nov 22 at 16:49
Paras Khosla
449
449
add a comment |
add a comment |
2 Answers
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votes
up vote
1
down vote
accepted
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 at 16:52
José Carlos Santos
147k22117217
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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 at 16:53
Nosrati
26.3k62353
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 at 16:52
José Carlos Santos
147k22117217
add a comment |
up vote
1
down vote
accepted
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 at 16:52
José Carlos Santos
147k22117217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 at 16:52
José Carlos Santos
147k22117217
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 at 16:52
José Carlos Santos
147k22117217
answered Nov 22 at 16:52
José Carlos Santos
147k22117217
answered Nov 22 at 16:52
José Carlos Santos
147k22117217
answered Nov 22 at 16:52
José Carlos Santos
147k22117217
147k22117217
add a comment |
add a comment |
up vote
1
down vote
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 at 16:53
Nosrati
26.3k62353
add a comment |
up vote
1
down vote
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 at 16:53
Nosrati
26.3k62353
add a comment |
up vote
1
down vote
up vote
1
down vote
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 at 16:53
Nosrati
26.3k62353
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 at 16:53
Nosrati
26.3k62353
answered Nov 22 at 16:53
Nosrati
26.3k62353
answered Nov 22 at 16:53
Nosrati
26.3k62353
answered Nov 22 at 16:53
Nosrati
26.3k62353
26.3k62353
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