1
$\begingroup$

This problem (if my derivations of them are correct) lead me to calculate the following integrals:

$$I_1 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\frac{-1+x_1x_2-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_2 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\frac{1-x_1x_2+2x_1-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_3 = \int_0^2 { \int_0^{\frac{1}{2}x_1} {\frac{-1-x_1x_2+2x_1-2x_2}{x_1-2x_2}} }dx_2dx_1$$ $$I_4 = \int_0^2 { \int_{\frac{1}{2}x_1}^1 {\frac{1+x_1x_2-2x_2}{x_1-2x_2}} }dx_2dx_1$$

How can this be done?

$\endgroup$
1
$\begingroup$

All integrals $I_n$, $n=1,2,3,4$, diverge logarithmically. Let's have a look at $I_1$. Here, $I_1=J_1+J_2+J_3$ where

$$J_1=\int_0^2\int_0^{x/2}\frac{1}{2y-x}dy\,dx$$

for which the inner integral

$$\int_0^{x/2}\frac{1}{2y-x}dy=(\frac12 \log|2y-x|)|_0^{x/2} \cdots \text{is undefined at the upper limit} $$ __________________________________________

$$J_2=\int_0^2\int_0^{x/2}\frac{-xy}{2y-x}dy\,dx$$

for which the inner integral

$$\begin{align} \int_0^{x/2}\frac{-xy}{2y-x}dy&=-\frac{x}{2} \int_0^{x/2}(1+\frac{x}{2y-x})dy\\\\ &=-\frac{x^2}{4}-\frac{x^2}{4}(\log|2y-x|)|_0^{x/2}\,\cdots \text{is undefined at the upper limit} \end{align}$$ __________________________________________

and

$$J_3=\int_0^2\int_0^{x/2}\frac{2y}{2y-x}dy\,dx$$

for which the inner integral diverges similarly.

$\endgroup$
  • $\begingroup$ I see my error now. I should have $\max(2, \cdot)$ and $\min(1, \cdot)$ in the integrand. $\endgroup$ – ploosu2 Apr 30 '15 at 19:29
  • $\begingroup$ Well, I guess there's no point in editing my question. I'll accept this answer for this question. But just to make sure: When you split the integrand, is this allowed, since you could also split $1 = \frac{x-y}{x-y}$? $\endgroup$ – ploosu2 Apr 30 '15 at 19:32
  • $\begingroup$ Well, technically it is not legitimate. I should have first written the upper limit as $x/2 - \epsilon$, where $\epsilon>0$. Then, we can split the integral, conduct the analysis, and take a limit as $\epsilon \to 0^+$. The conclusion is unaffected. $\endgroup$ – Mark Viola Apr 30 '15 at 19:35

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.