Please help me evaluate this seemingly simple integral

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In summary, the integral evaluates to 1 / (100 + iω) when x(t) is zero for t<0 and 1 / (100 + iω) when x(t) is 1 for t>0.
  • #1
tamtam402
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Homework Statement



Let x(t) = e-100tu(t)

u(t) = 0 for t < 0
u(t) = 1 for t > 0

Evaluate the following integral (from -∞ to ∞):

X(ω) = ∫ x(t)e-iωtdt

Homework Equations



See below.

The Attempt at a Solution



I tried to evaluate the integral by splitting it in two parts, since x(t) takes two different values. Keep in mind I'm replacing ∞ with a to evaluate the limit later.

from -∞ to 0:
X1(ω) = ∫ e-iωtdt

X1(ω) = (1/-iω) e-iωt (from -∞ to 0)

X1(ω) = (1/-iω)(1 - eiωa)


from 0 to ∞:
X2(ω) = e-100∫ e-iωtdt

X2(ω) = e-100 (1/-iω) e-iωt (from 0 to ∞)

X2(ω) = e-100 (1/-iω)(e-iωa - 1)


X(ω) = X1(ω) + X2(ω) = (1/-iω)(1 - eiωa) + e-100 (1/-iω)(e-iωa - 1)

= (1/-iω) (1 - eiωa + e-100e-iωa - e-100)


This is where I'm stuck. The answer is supposed to be X(ω) = 1 / (100 + iω), and I have absolutely no idea how I'm supposed to get there. Would someone mind helping me out?
 
Last edited:
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  • #2
Doh, it looks like I failed hard, I accidently put x(t) = e-100 instead of x(t) = e-100t. However when I tried to solve the problem earlier I got that right and I still didn't end up with the right answer (I re-did the problem from scratch when posting here). Is there a "trick" to solve this?
 
  • #3
shouldn't the negative half of the integral be zero?

also don't forget to take the limit for a as well
 
Last edited:
  • #4
lanedance said:
shouln;t thenegative half of the integral be zero?

also don't forget to take the limit for a as well

edit: sorry of some of my vocabulary is hard to understand, english isn't my primary language. I got the positive part of the integral right (it gives me the correct answer), I fixed the mistake I did in my first post. However I can't see how the negative part is supposed to be zero.

Well I see that it's supposed to be zero given the form of the answer, but I don't see how lim a -> ∞(1 - eiωa) can give zero, unless it's a special property of the exponent with the imaginary number i. Note that I'm taking the limit of a to infinity since I replaced the -∞ with -a earlier, thus cancelling the negative sign of the exponent.
 
  • #5
tamtam402 said:
edit: sorry of some of my vocabulary is hard to understand, english isn't my primary language. I got the positive part of the integral right (it gives me the correct answer), I fixed the mistake I did in my first post. However I can't see how the negative part is supposed to be zero.

Well I see that it's supposed to be zero given the form of the answer, but I don't see how lim a -> ∞(1 - eiωa) can give zero, unless it's a special property of the exponent with the imaginary number i. Note that I'm taking the limit of a to infinity since I replaced the -∞ with -a earlier, thus cancelling the negative sign of the exponent.

based on x(t)=0 for t<0, it should just be the integral of zero
 
  • #6
You have defined [itex]u(t)=0[/itex] when [itex]t<0[/itex] and [itex]x(t)=e^{-100t}\cdot u(t).[/itex]
 
  • #7
Yes, but I'm not evaluating the integral of x(t), I'm evaluating the integral of X(ω), which is different from x(t). Sorry if the notation is confusing, I used the variables given in the problem.
 
  • #8
tamtam402 said:
X(ω) = ∫ x(t)e-iωtdt
...
from -∞ to 0:
X1(ω) = ∫ e-iωtdt
...
Did you forget x(t) from the integrand?
 
  • #9
Wow I'm a dumbass, somehow I had x(t)=1 when t **< 0. Thanks everyone.
 
  • #10
No problem. I suspect that we have all made these kinds of mistakes, it often happens to me when the problem statement is complicated. :)
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve. It is used to determine the total value or quantity of a function over a given interval. In other words, it is a way to find the value of a function by adding up infinitely small parts.

2. Why is evaluating integrals important?

Evaluating integrals is important because it allows us to solve a wide range of mathematical problems in various fields, such as physics, economics, and engineering. It also helps us to understand the behavior of functions and calculate important quantities like displacement, velocity, and acceleration.

3. How do I evaluate an integral?

To evaluate an integral, you can use various methods such as substitution, integration by parts, or partial fractions. It is essential to understand the fundamental rules and properties of integrals, such as the power rule, the sum rule, and the constant rule, to correctly solve an integral.

4. What makes evaluating integrals challenging?

Evaluating integrals can be challenging because there are many different methods and techniques to solve them, and it can be difficult to determine which method is the most appropriate for a particular integral. Additionally, some integrals may not have an exact solution, and numerical methods may need to be used to approximate the integral.

5. Are there any tips for solving integrals more efficiently?

Yes, there are a few tips that can help you solve integrals more efficiently. Firstly, make sure you understand the fundamental rules and properties of integrals. Then, try to simplify the integrand by using algebraic manipulation or trigonometric identities. It is also helpful to practice and get familiar with different integration techniques to determine which one is the most suitable for a given integral.

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