- #1

hogrampage

- 108

- 1

[itex]\int^{\infty}_{0}2e^{2+jωt}dt[/itex] = [itex]\frac{j2e^{2}}{\omega}[/itex]

Why is it not ∞? Here are my steps:

Let u = 2+jωt, du = jωdt, dt = [itex]\frac{1}{jω}[/itex]du = -[itex]\frac{j}{ω}[/itex]du

[itex]\int^{\infty}_{0}2e^{2+jωt}dt[/itex]

= -[itex]\frac{2j}{ω}[/itex][itex]\int^{\infty}_{2}2e^{u}du[/itex]

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

= -[itex]\frac{2j}{ω}[/itex][itex]\stackrel{lim}{h\rightarrow∞}[/itex]([itex]e^{h}-e^{2}[/itex])

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