How can one derive the above equality?
Oh, there's a condition on $x$, which is $0
Any help would be very much appreciated!
How can one derive the above equality?
Oh, there's a condition on $x$, which is $0
Any help would be very much appreciated!
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 $