# Evaluate improper integral: Discontinuous integrand

1. Oct 14, 2014

### HugoAng

1. The problem statement, all variables and given/known data:

Evaluate: ∫-214 (1+X)-1/4

2. Relevant equations

ab f(x)dx =
lim ∫at f(x)dx
t→b-

And

ab f(x)dx =
lim ∫tb f(x)dx
t→a+

3. 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:

-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: Oct 14, 2014
2. Oct 14, 2014

### GFauxPas

Why is it a bad thing that you're getting a complex number as an answer?

3. Oct 14, 2014

### HugoAng

Well because as of now I do not what it means in regards to the concept. We have not studied that yet.

4. Oct 14, 2014

### Staff: Mentor

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. Oct 14, 2014

### Ray Vickson

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. Oct 14, 2014

### HugoAng

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]

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