How to calculate the integral of exp(-ikt) from -infinity to zero

In summary, the conversation discusses how to calculate the integral of e^(-ikx) from -∞ to 0. It is mentioned that the answer involves using complex analysis and making a substitution of k→k+iε. The speaker also provides hints on how to approach the problem, including factoring the exponential and using integration by parts. The final answer is expected to be similar to the guessed answer, but with some differences.
  • #1
PhilandSci
1
0

Homework Statement



Hi everyone. I am trying to calculate $$f(k) = \int_{-\infty}^0 e^{-i k t} dt.$$

Homework Equations



Of course we know that $$\int_{-\infty}^{\infty} e^{-i k t} dt = 2 \pi \delta(k)$$ and in fact (I think) I know that the answer to my question must be $$f(k) = \frac{i}{k} + \pi \delta(k).$$ But I am puzzled as to how exactly one can show that this is the answer.

The Attempt at a Solution



I assume we need to make a substitution $$t \rightarrow t + i\epsilon,$$ use complex analysis, and eventually let $$\epsilon \rightarrow 0.$$ But how exactly do we do that? What theorem in complex analysis do we use and what would the result be before epsilon is set to zero?
 
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  • #2
This is a nice but sort of tough question. Here are some hints for you.

First, doing t→t+iε doesn't seem to lead anywhere very helpful. However, you should try k→k+iε. To make it look nicer I changed variables to x=-t. You can factor the exponential in the integrand into two terms, one like eikx and another term like e-εx. Now try doing integration by parts. As with other trigonometric integrals, when you change it into the form uv-∫vdu, it turns out that second integral is related to the integral you want. You need to do a little algebra and be careful about taking limits, and you should get an answer similar to (but not exactly) what you guessed.
 

1. What is the integral of exp(-ikt) from -infinity to zero?

The integral of exp(-ikt) from -infinity to zero is equal to 1/ik, where i is the imaginary unit and k is a constant. This can be written as ∫exp(-ikt)dx = 1/ik + C, where C is the constant of integration.

2. How do you calculate the integral of exp(-ikt) from -infinity to zero?

To calculate the integral of exp(-ikt) from -infinity to zero, you can use the basic integration rule for exponential functions, which states that ∫exp(ax)dx = (1/a)exp(ax) + C. In this case, a = -ik, so the integral becomes ∫exp(-ikt)dx = (1/-ik)exp(-ikt) + C. After applying the constant of integration, the final answer is 1/ik + C.

3. Can the integral of exp(-ikt) from -infinity to zero be solved without using integration rules?

Yes, the integral of exp(-ikt) from -infinity to zero can also be solved using integration by parts. This method involves splitting the integral into two parts and using the formula ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)u'(x)dx, where u(x) and v(x) are functions and u'(x) and v'(x) are their derivatives. After applying this formula, the integral can be solved without using integration rules.

4. Is the integral of exp(-ikt) from -infinity to zero a definite or indefinite integral?

The integral of exp(-ikt) from -infinity to zero is a definite integral, as it has specific limits of integration (-infinity and zero) and will result in a numerical value. Conversely, an indefinite integral has no specific limits and will result in a function. In this case, the integral of exp(-ikt) from -infinity to zero results in the numerical value of 1/ik.

5. Can the integral of exp(-ikt) from -infinity to zero be solved using the substitution method?

Yes, the integral of exp(-ikt) from -infinity to zero can also be solved using the substitution method. This involves substituting a new variable, such as u = -ikt, and then replacing all instances of x with the new variable. After solving the integral with the new variable, the final answer can be converted back to the original variable, x. This method can be useful for more complex integrals involving trigonometric functions or nested expressions.

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