Solving an integral including a plus distribution.
Sophia Terry 
I am trying to solve \begin{equation}\int_0^1 \text{d}x \left(\frac{1+x^2}{1-x}\right)_+,\end{equation} where the plus should represent a plus distribution. For a plus distribution $T_{[f(x)]_+}$ I have the following: $$T_{[f(x)]_+}[\varphi]\equiv \int_0^1 f(x)[\varphi(x)-\varphi(1)]\text{d}x.$$ Now I am not quite sure how to understand the notation of the integral I am trying to solve. Furthermore, it seems that this integral would diverge. However, I tried to solve it with the substitution $x=u-1$ and came up with the following: \begin{align} \int_0^1 \text{d}x \frac{1+x^2}{1-x}&=\int_1^2 \frac{1+(u-1)^2}{1-u+1}\text{d}u\\ &=\int_1^2\frac{2+u^2-2u}{-u}\text{d}u\\ &=-2\log u -\frac{u^2}{2}+2u\bigg|_1^2\\ &=-\log 4+\frac{1}{2}. \end{align} Does this make sense? How else can I interpret the notation of this integral?
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