Integration Woes!!!!

[mathNewbie]

Hello! I am having difficulty solving a double integral problem. The problem is stated as such (I apologize that I do not know how to type integration signs):

Integral(from 0 to 1) of Integral (from 0 to 1) of dxdy/(1-xy).

i know that the integral of 1/x is ln(x), but how do i go about finding the integral of 1/1-xy. I know i treat y as a constant when integrating with respect to x, yes? so i can think of the integral temporarily as 1/(1-kx) where k is some constant y? i think this is right but i still don't know how to solve the integral... can anyone point me in the right direction?

Thanks!

 

[pappus]

Hello! I am having difficulty solving a double integral problem. The problem is stated as such (I apologize that I do not know how to type integration signs):

Integral(from 0 to 1) of Integral (from 0 to 1) of dxdy/(1-xy).

i know that the integral of 1/x is ln(x), but how do i go about finding the integral of 1/1-xy. I know i treat y as a constant when integrating with respect to x, yes? so i can think of the integral temporarily as 1/(1-kx) where k is some constant y? i think this is right but i still don't know how to solve the integral... can anyone point me in the right direction?

Thanks!

1. Evaluate the integral in two steps:

\(\displaystyle \int_0^1 \underbrace{\int_0^1\left(\frac1{1-xy} \right)dx}_{\text{do this first. y is a constant}} dy\)

\(\displaystyle \int_0^1\left(\frac1{1-xy} \right)dx = \left[ -\ln(1-xy) \right]_0^1 = -\ln(1-y)\)

2. Your integral is now:

\(\displaystyle \int_0^1\left(-\ln(1-y) \right)dy\)

Use integration by parts.

 

[HallsofIvy]

That's not quite right. There is a "y" missing.

\(\displaystyle \int_0^1 \frac{1}{1- xy}dx\)
Let u= 1- xy so that du= -y dx, dx=-(1/y)du, dx. When x= 0, u= 1, when x= 1, u= 1- y so the integral becomes
\(\displaystyle - \frac{y}\int_1^{1- y}\frac{1}{u}du= -\frac{(ln(1- y)}{y}\)
so \(\displaystyle \int_0^1\int_0^1 \frac{1}{1- xy}dxdy= -\int_0^1 \frax{ln(1- y)}{y}dy\).

 

[mathNewbie]

thanks so much for the help!! but how do i go about solving the next integral? do i do u substitution again?

 

[srmichael]

thanks so much for the help!! but how do i go about solving the next integral? do i do u substitution again?

No. Use integration by parts like pappus suggested. Have you learned that method of integration?

 

[mathNewbie]

unfortunately i've never heard of it. i know that if i have the integral of x + 2x i can just do the integral of x plus the integral of 2x right? is it something along those lines?

 

[srmichael]

unfortunately i've never heard of it. i know that if i have the integral of x + 2x i can just do the integral of x plus the integral of 2x right? is it something along those lines?

No, it's not. Click on this link (integration by parts) to see what it's all about.