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

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In summary, the integral \int^{\infty}_{0}2e^{2+jωt}dt does not converge for all values of ω. It only converges for ω < 0, and the correct answer for this integral is \frac{j2e^{2}}{ω}.
  • #1
hogrampage
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I do not understand the following integral:

[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]?
 
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  • #2
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.
 
  • #3
hogrampage said:
I do not understand the following integral:

[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]?

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

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

Now, for what values of a and b will that expression converge?
 
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  • #4
jackmell said:
That's cus you don't know what [itex]\omega[/itex] is. Let's just look at:
[tex]\int_0^{\infty} e^{iwt}dt=\frac{1}{iw}e^{iwt}\biggr|_0^{\infty}[/tex]

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

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

a, b < 0 which means ω < 0. So:

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

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

[itex]-\frac{2e^{2}}{jω} = \frac{j2e^{2}}{ω}[/itex] for ω < 0.

Thanks for the help! If I made any mistakes above, let me know.
 
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  • #5
hogrampage said:
a, b < 0 which means ω < 0. So:

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

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

[itex]-\frac{2e^{2}}{jω} = \frac{j2e^{2}}{ω}[/itex] 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.
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value or quantity of a function over a given interval.

2. What is the purpose of solving an integral?

Solving an integral allows us to find the exact value of a function over a given interval. This is useful in many areas of science and engineering where we need to calculate the total value of a quantity or the area under a curve.

3. What does the notation 2exp(2+jωt) mean?

The notation 2exp(2+jωt) means 2 multiplied by the exponential function e raised to the power of 2+jωt. This is a common way to represent complex numbers in mathematical equations.

4. How do you solve an integral with a complex number?

To solve an integral with a complex number, we can use the same techniques as solving a regular integral. However, we need to be careful when dealing with the complex number and follow the rules of complex arithmetic. In this case, we would need to use the properties of the exponential function to simplify the integral.

5. Can you explain the answer to Solving Integral 2exp(2+jωt)?

The answer to Solving Integral 2exp(2+jωt) is the integral of 2 multiplied by the exponential function e raised to the power of 2+jωt. Depending on the limits of integration, the answer may vary. To solve this integral, we would need to use the properties of the exponential function and complex arithmetic to simplify the expression and then integrate it using standard techniques.

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