# I Non-Integrable Function

Tags:
1. Mar 21, 2017

### junt

Hi All,

I am wondering if the function below is Integrable:

$$\frac{\exp{(-\frac{1}{2}(u-2)^2-2u^2)}}{u-2}$$

When I work it out on computer, the integral is finite from -Inf to Inf. But clearly it has a pole at u=2. Is this pole integrable? If yes, what kind of coordinate transform is required?

Any help is appreciated!

Last edited: Mar 21, 2017
2. Mar 22, 2017

### haruspex

in the vicinity of u=2, the numerator is bounded away from zero. The integral from 2-ε to 2 is -∞, while that from 2 to 2+ε is +∞. Trying to integrate through u=2 might effectively cancel these, leading to an apparently valid result, but it is not valid. -∞+∞ is undefined.

3. Mar 22, 2017

### aheight

It is integrable in the Cauchy Principal-Value sense: We take limits:

$\displaystyle\lim_{\epsilon\to 0} \big\{\int_{-\infty}^{2-\epsilon} f(u)du+\int_{2+\epsilon}^{\infty}f(u)du\big\}$

We can numerically compute it in Mathematica:
Code (Text):
In[104]:=
NIntegrate[Exp[(-2^(-1))*(u - 2)^2 - 2*u^2]/(u - 2), {u, -Infinity, 2, Infinity}, Method -> PrincipalValue]

Out[104]=
-0.15723961536751363

4. Mar 22, 2017

### junt

That looks totally good. But when I modify my Integrand with different parameters, the same Integral doesn't seem to be Integrable under Cauchy Principal-Value. For instance:

$$2\frac{\exp{(-\frac{1}{4}(-2+u)^2-u^2)}}{|u-2|}$$

This integral doesn't converge in mathematica. Why is it so?

Code (Text):
In[104]:=
NIntegrate[2 Exp[-1/4*(u - 2)^2 - u^2]/Abs[(u - 2)], {u, -Infinity, 2, Infinity}, Method -> PrincipalValue]

Says failed to converge. Isn't it the same type of integral as my original post. I just changed some parameters.

One more thing, I originally had forgotten the absolute value on the denominator. So now, it looks like even my original integrand doesn't converge when I force my denominator to be positive.

Last edited: Mar 22, 2017
5. Mar 22, 2017

### haruspex

Sure, but I find that very unconvincing. If you change one of the limits to 2-2ε it doesn't converge (does it?)

6. Mar 22, 2017

### FactChecker

It's related to the contour integral of the analytic function and its residue, which is much less arbitrary.

7. Mar 23, 2017

### aheight

First, realize we are basically looking at a variant of $\frac{1}{u}$. Basically. And we can integrate this in the principal value sense:

$\text{P.V.} \int_{-\infty}^{\infty} 1/u du=0$

That's because the infinities cancel when taking "symmetric limits" ( as $\epsilon\to 0$). However, when we take the absolute values, the infinities DO NOT cancel.

Edit: Maybe not exactly a variant of 1/u but I think the two are similar qualitatively and looking at 1/u is easier to visualize and easier to understand.

Last edited: Mar 23, 2017