# Integrate 2exp(2+jωt)

hogrampage
I do not understand the following integral:

$\int^{\infty}_{0}2e^{2+jωt}dt$ = $\frac{j2e^{2}}{\omega}$

Why is it not ∞? Here are my steps:

Let u = 2+jωt, du = jωdt, dt = $\frac{1}{jω}$du = -$\frac{j}{ω}$du

$\int^{\infty}_{0}2e^{2+jωt}dt$

= -$\frac{2j}{ω}$$\int^{\infty}_{2}2e^{u}du$

= -$\frac{2j}{ω}$$\stackrel{lim}{h\rightarrow∞}$$\int^{h}_{2}2e^{u}du$

= -$\frac{2j}{ω}$$\stackrel{lim}{h\rightarrow∞}$($e^{h}-e^{2}$)

To me, this limit does not exist, so why is the answer $\frac{j2e^{2}}{\omega}$?

Homework Helper
You also need to change your integration limits, if ##t \to \infty## then ##u \to 2 + j \infty##.

First thing I would do is separate ##e^{2 + j \omega t} = e^2 e^{j \omega t}## and pull the e² in front of the integration sign. The remaining integral can be solved using for example contour integration.

jackmell
I do not understand the following integral:

$\int^{\infty}_{0}2e^{2+jωt}dt$ = $\frac{j2e^{2}}{\omega}$

Why is it not ∞? Here are my steps:

Let u = 2+jωt, du = jωdt, dt = $\frac{1}{jω}$du = -$\frac{j}{ω}$du

$\int^{\infty}_{0}2e^{2+jωt}dt$

= -$\frac{2j}{ω}$$\int^{\infty}_{2}2e^{u}du$

= -$\frac{2j}{ω}$$\stackrel{lim}{h\rightarrow∞}$$\int^{h}_{2}2e^{u}du$

= -$\frac{2j}{ω}$$\stackrel{lim}{h\rightarrow∞}$($e^{h}-e^{2}$)

To me, this limit does not exist, so why is the answer $\frac{j2e^{2}}{\omega}$?

That's cus you don't know what $\omega$ is. Let's just look at:
$$\int_0^{\infty} e^{iwt}dt=\frac{1}{iw}e^{iwt}\biggr|_0^{\infty}$$

Let $\omega=a+bi$ then we have
$$\frac{1}{i(a+bi)}e^{i(a+bi)t}\biggr|_0^{\infty}$$

Now, for what values of a and b will that expression converge?

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hogrampage
That's cus you don't know what $\omega$ is. Let's just look at:
$$\int_0^{\infty} e^{iwt}dt=\frac{1}{iw}e^{iwt}\biggr|_0^{\infty}$$

Let $\omega=a+bi$ then we have
$$\frac{1}{i(a+bi)}e^{i(a+bi)t}\biggr|_0^{\infty}$$

Now, for what values of a and b will that expression converge?

a, b < 0 which means ω < 0. So:

$\frac{1}{iw}e^{iwt}\biggr|_0^{\infty} = -\frac{1}{iω}$ for ω < 0.

From that, the complete answer to the original integral is:

$-\frac{2e^{2}}{jω} = \frac{j2e^{2}}{ω}$ for ω < 0.

Thanks for the help! If I made any mistakes above, let me know.

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jackmell
a, b < 0 which means ω < 0. So:

$\frac{1}{iw}e^{iwt}\biggr|_0^{\infty} = -\frac{1}{iω}$ for ω < 0.

From that, the complete answer to the original integral is:

$-\frac{2e^{2}}{jω} = \frac{j2e^{2}}{ω}$ for ω < 0.

Thanks for the help! If I made any mistakes above, let me know.

Your analysis of a and b is not correct. for w=a+bi, in order for the integral to converge, b has to be less than zero. a can be any real number. Go through that to make sure you understand it.