2
$\begingroup$

Cauchy principal value of the following integration gives nothing in Mathematica.

$\int\limits_{-\infty}^\infty\frac{\text{dx}}{x-2}$

I tried some hand calculation which is

$\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\int\limits_{-R}^{2-\epsilon}\frac{\text{dx}}{x-2}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\int\limits_{2+\epsilon}^{R}\frac{\text{dx}}{x-2}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\left.\ln(x-2)\right\rvert_{-R}^{2-\epsilon}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\left.\ln(x-2)\right\rvert_{2+\epsilon}^{R}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\ln\frac{2-\epsilon-2}{-R-2}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\frac{R-2}{2+\epsilon-2}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\ln\left(\frac{-\epsilon}{-R-2}\times\frac{R-2}{\epsilon}\right)$

$=\lim\limits_{R\rightarrow\infty}\ln\left(\frac{R-2}{R+2}\right)$

$=\lim\limits_{R\rightarrow\infty}\ln\left(\frac{1-2/R}{1+2/R}\right)$

$=\ln(1)=0$

I used the following command

Integrate[1/(x - 2), {x, -Infinity, Infinity}, PrincipalValue -> True]

which returns the same.

However,

Integrate[1/((x - 2) (x - 3)), {x, -Infinity, Infinity}, PrincipalValue -> True]

returns zero. Why the first one fails to compute the result? (Or am I doing anything wrong in the calculation?)

$\endgroup$
  • 4
    $\begingroup$ First one fails due to divergence at +-infinity, which are not first order poles for the function. $\endgroup$ – Daniel Lichtblau May 4 '16 at 17:28
  • $\begingroup$ @DanielLichtblau But, any antisymmetric integrand integrated over the real line must integrate to zero. (There is no difference between 1/(x - 2) and 1/x in this regard.) $\endgroup$ – bbgodfrey May 6 '16 at 0:02
  • $\begingroup$ @ bbgodfrey So Integrate[1/x, {x, -Infinity, Infinity}, PrincipalValue -> True] claims divergence, for eactly the same reason. $\endgroup$ – Daniel Lichtblau May 6 '16 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.