WHY is this integral divergent? (another question)

**s3a** · March 30th 2010, 09:36 AM

Why is this integral divergent?:

I get 1/2 * (ln 1 - ln(inf) + ln(inf) - ln 1) so I thought the answer would be zero. Or do we assume that the infinities are different in each ln therefore it diverges? Or am I completely off?

Any input would be greatly appreciated!
Thanks in advance!

**Black** · March 30th 2010, 10:23 AM

Split the integral up

$\int_{-\infty}^{\infty}\frac{x}{x^2+11} \, dx=\int_{-\infty}^{0}\frac{x}{x^2+11} \,dx+\int_{0}^{\infty}\frac{x}{x^2+11} \, dx$ .

Let's do the second integral on the right.

$\int_{0}^{\infty}\frac{x}{x^2+11} \, dx= \frac{1}{2} \lim_{t \to \infty} \left[ \ln(x^2+11)\right]_{0}^t= \infty$ .

Therefore, the integral $\int_{-\infty}^{\infty}\frac{x}{x^2+11} \, dx$ is divergent.

**HallsofIvy** · March 30th 2010, 01:23 PM

[/tex]lim_{a\to\infty}\int_{-a}^a f(x)dx[/tex] is often called the "Cauchy Principal Value" of $\int_{-\infty}^\infty f(x) dx$ and often exists when the integral itself, which is defined as $\lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b f(x) dx$ does not exist.

Here, the integral intself, where the two limits must be taken independently does not exist- what you calculated is the "Cauchy Principal Value".

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