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How can one derive the above equality?

Oh, there's a condition on $x$, which is $0

Any help would be very much appreciated!

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    I think the first step should be $\int_x^1\int_x^1\frac{dsdt}{1-st}=2\int_x^1(\int_t^1\frac{ds}{1-st})dt$. Actually, I think I already know how to derive this identity. Anyway, it's a good exercise, so I'll leave other people write the whole solution.2012-04-07

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This is extended version of the caozhu's comment.

Note that integration goes over square $[x,1]\times[x,1]$ and function $f(s,t)=\frac{1}{1-st}$ is symmetric, i.e. $f(s,t)=f(t,s)$ hence $ \int\limits_x^1\int\limits_t^1f(s,t)dsdt= \int\limits_x^1\int\limits_s^1f(s,t)dtds $ So, $ \int\limits_x^1\int\limits_x^1f(s,t)dsdt= \int\limits_x^1\int\limits_t^1f(s,t)dtds+ \int\limits_x^1\int\limits_s^1f(s,t)dsdt= 2\int\limits_x^1\int\limits_t^1f(s,t)dtds $ Then we get iterated integral $ \int\limits_x^1\int\limits_x^1f(s,t)dsdt= 2\int\limits_x^1dt\int\limits_t^1\frac{ds}{1-st}= 2\int\limits_x^1dt\left(-\frac{\log(1-st)}{t}\right)_{s=t}^{s=1}= 2\int\limits_x^1\frac{\log(1+t)}{t}dt $