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I came across a question that states that the following integral does not exist

$$\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx$$

I have seen similar questions on here where people have said that they can be integrated and have shown this. I am confused as to whether it can be or not.

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    $\begingroup$ Split the integral at $x=-2$. At best the lower integral is $-\infty$ and the higher integral $+\infty$ making the sum $\infty-\infty$, which is not well defined $\endgroup$ – Henry Jun 25 at 7:40
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    $\begingroup$ Could you link to some of these questions? Context is important with questions like these. $\endgroup$ – Theo Bendit Jun 25 at 7:41
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    $\begingroup$ What happens around $-2$? $\endgroup$ – badjohn Jun 25 at 7:41
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It depends on what type of integral you are considering. For example the integral from $-4$ to $-2-\epsilon$ plus the integral from $-2+\epsilon $ to $0$ vanishes for every $\epsilon >0$ so the given integral is $\int_0^{4} \frac 1{x+2} dx=ln(6)-ln(2)=\ln(3)$ in a limiting sense. But the integral from $-2$ to $4$ is $\infty$ by direct calculation , so in the usual sense the integral does not exist.

I am thankful to Raymond Manzoni for his help in correcting a mistake.

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  • $\begingroup$ Integral from $-4$ to $4$ is not equal to anything in the Riemann or Lebesgue sense. But integral from $-2$ to $4$ is certainly $+\infty$. $\endgroup$ – Kavi Rama Murthy Jun 25 at 8:41
  • $\begingroup$ @RaymondManzoni I am really grateful to you for your comments. $\endgroup$ – Kavi Rama Murthy Jun 25 at 8:42
  • $\begingroup$ you are taking the same $\epsilon$ on both sides, and then it disappears. But how to support that $\epsilon$ be the same ? $\endgroup$ – G Cab Jun 25 at 10:26
  • $\begingroup$ That is how principal values of integrals are defined. For example $\int_{-1}^{1} \frac 1 x dx$ has principal value $0$ even though $\frac 1 x$ is not integrable. $\endgroup$ – Kavi Rama Murthy Jun 25 at 10:29
  • $\begingroup$ $0$ is a singularity for the function $\ln$ but $\displaystyle \int_0^1 \ln x\,dx=-1$ in an usual sense. $\endgroup$ – FDP Jun 25 at 12:01
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Strictly something like this cannot be integrated, as this singularity causes several problems. If we were to just calculate the value using normal calculus and ignore the presence of this we would find the Cauchy principle value, however this effectively assumes that the asymptote upwards is equal in magnitude to the one downwards, and so they 'cancel'. However, in most areas of maths you cannot assume "all infinities are equal" and so this would be seen as divergent.

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The Riemann integral $\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx$ as no defined value since the function $\frac{1}{x+2}$ is not finite at $x=-2$ and this point is in the range of integration.

But the integral has a different meaning in sens of Cauchy Principal Value :

$$PV\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx = \lim_{\epsilon\to 0}\left(\displaystyle \int_{-4}^{-2-\epsilon} \frac{1}{x+2} \,\mathrm dx +\displaystyle \int_{-2+\epsilon}^{4} \frac{1}{x+2} \,\mathrm dx\right)= \ln(3)$$

http://mathworld.wolfram.com/CauchyPrincipalValue.html

So take carre that the answer is not the same if the question refers to the value of a definite Riemann integral and an integral in sens of Cauchy Principal value which is denoted with $PV$ in front of the integral.

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