# My integral is behaving in strange way

The following is an integral whose behavior I cant understand

$$I=\int_{\gamma}^{\infty} \Bigg[ 1- 2a - \left( \frac{b}{1+s x^{-1} } \right) \Bigg] x^{-1/3} \, dx$$

with $$2a+b=1$$

Consider Case1: $$a=1/2$$ $$b=0$$ then the integral clearly is zero.

Now consider the case when: $$a=0.485$$ $$b=0.03$$ that is, I chose very close numbers to case 1 so as to understand its relationship with it.

Do you believe that the performance of both integrals should be comparable or very close to one another?

The reason I ask is this: I run some numerical simulations of expressions containing these two different cases and they don't give the comparable results. I evaluate $$\text{exp}[- N \ I]$$ where $N=130690$ to be precise.

I want to understand whether its a numerical error or these two are really different?

P.S A further note, after the answer I received, the integral is non divergent for Case 2.

Thanks.

• At what value is $\gamma$ and $s$? – flawr Feb 4 '15 at 21:48
• In the problem I solve, both $\gamma$ and $s$ are non-negative. The answer of this integral should be function of $s$. I didnt include the whole problem for sake of simplicity. – Tyrone Feb 4 '15 at 21:53

I just tried to simplify your integrand:

$$\Bigg[ 1- 2a - \left( \frac{b}{1+s x^{-1} } \right) \Bigg] x^{-1/3}$$

$$=\left[1-2a-\frac{xb}{x+s}\right]x^{-1/3}$$

$$=\left[1-(2a+b)+\frac{sb}{x+s}\right]x^{-1/3}$$

$$=\frac{sb}{x^{1/3}(x+s)}$$

And wolframalpha says if you integrate this from $0$ to $\infty$ you get

$$\frac{2 \pi sb}{\sqrt{3} s^{1/3}}$$

So there should not be any problem of convergence I think.

• thanks, do you believe with such a simplification I can understand the behavior ? – Tyrone Feb 4 '15 at 22:03
• Yes, this function looks as if it was integrable, as it converges faster to zero than e.g. $1/x$, since $1/x^a$ is integrable (on the interval $[1,\infty])$ for all $a>1$. – flawr Feb 4 '15 at 22:07
• Yes, but I dont understand why the two integrals are not almost equal. – Tyrone Feb 4 '15 at 22:08
• There is no problem with convergence, please check my other question: math.stackexchange.com/questions/1118345/… – Tyrone Feb 4 '15 at 22:09
• Well they are if you apply this formula I just provided! I do not see why you just don't use the indefinite integral you can explicitly calculate!!! – flawr Feb 4 '15 at 22:09

Your integral is an integral that runs to infinity. In this cases integrals can be divergent (i.e. the result of the integral is plus or minus infinity). When you try to apply the linearity rule for integrals and then try to integrate $\int_\gamma^\infty x^{-\frac{1}{3}}dx$ you will see why there was such a great difference between Case 1 and Case 2.

These divergences can be fixed with some mathematical tricks. For example you can replace $\infty$ by a variable $N$ and treat $N$ as a finite real number at first and after Integration you can set $N \rightarrow \infty$

• Case 2 is not a divergent integral for the choice of $a$ and $b$ provided. – Tyrone Feb 4 '15 at 22:01
• I have added an update to my answer. – kryomaxim Feb 4 '15 at 22:03
• But I don't have a problem with divergence... – Tyrone Feb 4 '15 at 22:04

When $b \neq 0$, and $2a + b = 1$ $$[1 - 2a - \frac{b}{1+sx^{-1}}]x^{-\frac13} = \frac{(1-2a)(x+s) - xb}{x + s}x^{-\frac13} = bx^{-\frac13}\frac{x+s - x}{x + s} = bs\frac{x^{-\frac13}}{x+s}$$

Assuming positive $\gamma, s$ this should converge. Further, $$\int_\gamma^{\infty} [1 - 2a - \frac{b}{1+sx^{-1}}]x^{-\frac13} dx = b\int_\gamma^{\infty} s\frac{x^{-\frac13}}{x+s}dx$$

So the value of your integral is in fact directly proportionate to $b$