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I was using a double integral to check for some constants.

I came across this one.

How can we show that

$$\int_{0}^{1}\int_{0}^{1}{1-x\over 1-xy}\cdot{x\over \ln{(xy)}}dxdy={1\over 2}\ln{1\over 2}$$

My try:

$$\int_{0}^{1}\int_{0}^{1}\left[{x(1-xy)^{-1}\over \ln{(xy)}}-{x^2(1-xy)^{-1}\over \ln{(xy)}}\right]dxdy$$

Apply binomial series: $$\int_{0}^{1}\int_{0}^{1}\left[{x-x^2y+x^3y^2-x^4y^3+\cdots\over \ln{(xy)}}-{x^2-x^3y+x^4y^2-x^5y^3+\cdots\over \ln{(xy)}}\right]dxdy$$

I wonder if we can apply the Frullani theorem at this point?

gymbvghjkgkjkhgfkl
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1 Answers1

7

Let $xy=u$ and $x=v$ so that your integral becomes, with Jacobian $-1/v$, $$-\int_0^1\int_0^v\frac{1-v}{(1-u)\ln u}dudv$$ which you can solve by changing the oreder of integration as $$-\int_0^1\int_u^1\frac{1-v}{(1-u)\ln u}dvdu=-\frac12\int_0^1\frac{1-u}{\ln u}du=\frac12\ln 2$$

Math-fun
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