1

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?

Community
  • 1
ploosu2
  • 8,707

1 Answers1

1

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.

Mark Viola
  • 179,405
  • I see my error now. I should have $\max(2, \cdot)$ and $\min(1, \cdot)$ in the integrand. – ploosu2 Apr 30 '15 at 19:29
  • 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 $1 = \frac{x-y}{x-y}$? – ploosu2 Apr 30 '15 at 19:32
  • 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. – Mark Viola Apr 30 '15 at 19:35