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integral - QuestionCove
OpenStudy (ksaimouli):

integral

1 year ago
OpenStudy (ksaimouli):

\[\int\limits_{0}^{} \frac{ \cos(2x) dx }{ 9x^2+4 }\]

1 year ago
OpenStudy (ksaimouli):

from 0 to infinity *{infinity is not printed on the integral}*

1 year ago
OpenStudy (ksaimouli):

\[\int\limits_{C}^{}\frac{ e^{i2z} }{ (z+2/3i)(z-2/3i) } dz\]

1 year ago
OpenStudy (ksaimouli):

\[\lim_{z \rightarrow 2/3i}\int\limits_{C}^{}\frac{ e^{i2z} }{ (z+2/3i) }\]

1 year ago
OpenStudy (ksaimouli):

I get \[\frac{ 3 \pi e^{-4/3} }{ 2}\], but the answer is different

1 year ago
OpenStudy (ksaimouli):

\[\lim_{z \rightarrow 2/3i}\frac{ e^{i2x} }{ z+(2/3i) }*2*\pi*i\]

1 year ago
OpenStudy (ksaimouli):

^ step 2 in detail

1 year ago
myininaya (myininaya):

i think your integral is just a tad off

1 year ago
OpenStudy (michele_laino):

we have this: \[\Large \int_0^{ + \infty } {} = \frac{1}{2}\int_{ - \infty }^{ + \infty } {} \]

1 year ago
myininaya (myininaya):

\[\int\limits\limits_0^\infty \frac{e^{2ix}+e^{-2x i}}{ 2} \cdot \frac{1}{(3x+2i)(3x-2i)} dx \\ =\frac{1}{18} \int\limits\limits_0^\infty \frac{e^{2i x}-e^{-2ix}}{(x+\frac{2i}{3})(x-\frac{2i}{3})} dx\]

1 year ago
myininaya (myininaya):

so if you multiply that result you got by 1/18 you will get wolfram's answer

1 year ago
OpenStudy (michele_laino):

I agree with @myininaya we have to make this substitution: \[\Large \cos \left( {2x} \right) = \frac{{{e^{i2x}} + {e^{ - i2x}}}}{2}\] then we have to go to the complex plane, and we have to apply the theorem of Jordan twice: |dw:1455658401988:dw|

1 year ago
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