Evaluate improper integral: Discontinuous integrand

In summary, the problem is that when you try to integrate the function ##f## from the interval ##(-2, -1)## to the interval ##(1, 14)##, the result is always complex, even when the original function is real. This is because the derivative of the function ##f## at ##(-2, -1)## is ##(-1)^{-1/4}(|x|-1)^{-1/4}##, and the derivative of the function at ##(1, 14)## is ##4/3(15)3/4##.
  • #1
HugoAng
3
0

Homework Statement

:

Evaluate: ∫-214 (1+X)-1/4[/B]

Homework Equations



ab f(x)dx =
lim ∫at f(x)dx
t→b-

And

ab f(x)dx =
lim ∫tb f(x)dx
t→a+

The Attempt at a Solution


So far what I have done is:
(-2,-1)∪(-1,14)

Thus I set up two integrals and chose one to evaluate: [/B]
-2-1 dx/((1+x)1/4 + ∫-114 dx/(1+x)1/4

The second integral gave me a finite answer of 4/3(15)3/4 but the first one always leads me to an imaginary unit. From this point I understand WHY this is happening, and that led me to think the problem may have a typo but the thing is my professor told us that there is an answer. Also, no one in my class got this question correct so it was turned into homework. Any help will be appreciated. Thanks!
 
Last edited:
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  • #2
Why is it a bad thing that you're getting a complex number as an answer?
 
  • #3
GFauxPas said:
Why is it a bad thing that you're getting a complex number as an answer?
Well because as of now I do not what it means in regards to the concept. We have not studied that yet.
 
  • #4
HugoAng said:

Homework Statement

:
[/B]
Evaluate: ∫-214 (1+X)-1/4

The Attempt at a Solution


So far what I have done is:
(-2,-1)∪(-1,14)

Thus I set up two integrals and chose one to evaluate: [/B]
-2-1 dx/((1+x)-1/4 + ∫-114 dx/(1+x)-1/4

The second integral gave me a finite answer of 4/3(15)3/4 but the first one always leads me to an imaginary unit. From this point I understand WHY this is happening, and that led me to think the problem may have a typo but the thing is my professor told us that there is an answer. Also, no one in my class got this question correct so it was turned into homework. Any help will be appreciated. Thanks!
I'm confused as to what the integral is. In the problem statement you have this:
$$\int_{-2}^{14} (1 + x)^{-1/4}dx$$
but in your work below there, you have integrands that look like this:
$$\frac{dx}{(1 + x)^{-1/4}}$$
 
  • #5
HugoAng said:

Homework Statement

:

Evaluate: ∫-214 (1+X)-1/4[/B]

Homework Equations



ab f(x)dx =
lim ∫at f(x)dx
t→b-

And

ab f(x)dx =
lim ∫tb f(x)dx
t→a+

The Attempt at a Solution


So far what I have done is:
(-2,-1)∪(-1,14)

Thus I set up two integrals and chose one to evaluate: [/B]
-2-1 dx/((1+x)-1/4 + ∫-114 dx/(1+x)-1/4

The second integral gave me a finite answer of 4/3(15)3/4 but the first one always leads me to an imaginary unit. From this point I understand WHY this is happening, and that led me to think the problem may have a typo but the thing is my professor told us that there is an answer. Also, no one in my class got this question correct so it was turned into homework. Any help will be appreciated. Thanks!

For ##x < -1## (with ##|x| > 1##) the integrand ##f(x) = (1+x)^{-1/4}## equals##(-1)^{-1/4}(|x|-1)^{-1/4}##. We have ##(-1)^{-1/4} = e^{-i \pi/4} = (1-i)/\sqrt{2}##, where ##i = \sqrt{-1}## is the imaginary unit. Since ##f## is complex-valued for ##x < -1## its integral from -2 to -1 is also complex-valued.
 
  • #6
Mark44 said:
I'm confused as to what the integral is. In the problem statement you have this:
$$\int_{-2}^{14} (1 + x)^{-1/4}dx$$
but in your work below there, you have integrands that look like this:
$$\frac{dx}{(1 + x)^{-1/4}}$$
Oops! Let me fix that real quick! I'm barely getting used to the forums.
It should be:
$$\frac{dx}{(1 + x)^{1/4}}$$[/QUOTE]
 

Related to Evaluate improper integral: Discontinuous integrand

1. What does it mean for an integral to have a discontinuous integrand?

A discontinuous integrand is one that is not continuous over the entire interval of integration. This means that the function being integrated has a break or jump in its graph, which can cause issues when trying to calculate the integral.

2. How do you evaluate an improper integral with a discontinuous integrand?

To evaluate an improper integral with a discontinuous integrand, the integral must be broken into smaller, continuous intervals. The limits of integration may also need to be adjusted to account for any breaks or jumps in the integrand.

3. Can an improper integral with a discontinuous integrand still have a finite value?

Yes, it is possible for an improper integral with a discontinuous integrand to have a finite value. This is especially true if the function has a vertical asymptote, as the integral can still converge to a finite value on either side of the asymptote.

4. How do discontinuities affect the convergence of an improper integral?

Discontinuities can affect the convergence of an improper integral by changing the behavior of the function being integrated. In some cases, the integral may no longer converge due to the break or jump in the integrand.

5. Are there any special techniques for evaluating improper integrals with discontinuous integrands?

Yes, there are special techniques that can be used to evaluate improper integrals with discontinuous integrands. These techniques include using limits, splitting the integral into smaller intervals, and applying the properties of integrals such as linearity and substitution.

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