Solving Integral 2exp(2+jωt): Explaining Answer

[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}?

 

[CompuChip]

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?

 

[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.

 

[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.