I'm trying to solve the following integral \begin{equation} I=\int_{0}^{\infty}e^{-x^2}\ln(x)\mathrm{d}x\;\text{.} \end{equation} After some time I found out that \begin{equation} I=\frac{1}{4}\Gamma'\left(\frac{1}{2}\right) \end{equation} Then I tried to evaluate $\Gamma'\left(\frac{1}{2}\right)$ which lead me to \begin{equation} \Gamma'\left(\frac{1}{2}\right)=-\sqrt{\pi}\left(2+\gamma+\sum_{n=1}^{\infty}\frac{2}{1+2n}-\frac{1}{n}\right) \end{equation} After that I tried to evaluate the sum with some integrals. Therefore \begin{equation} \sum_{n=1}^{\infty}\frac{2}{1+2n}-\frac{1}{n}=\sum_{n=1}^{\infty}\left(2\int_{0}^{1}x^{2n}\mathrm{d}x-\int_{0}^{1}x^{n-1}\mathrm{d}x\right) \end{equation} After some change of index I finally came to \begin{equation} \sum_{n=0}^{\infty}\left(2\int_{0}^{1}x^{2n+2}\mathrm{d}x-\int_{0}^{1}x^{n}\mathrm{d}x\right) =2\int_{0}^{1}\frac{x^2}{1-x^2}\mathrm{d}x-\int_{0}^{1}\frac{1}{1-x}\mathrm{d}x \end{equation}
Now my problem:
In the last line is a mistake and I don't know why. Before I used the geometric series the integral numerically goes to $-2+\ln(4)$, but after the use it goes to $-2+\ln(2)$. Why is here an interchange of limits not justified?
\begin{align} \sum_{n=0}^{\infty}\left(2\int_{0}^{1}x^{2n+2}\mathrm{d}x-\int_{0}^{1}x^{n}\mathrm{d}x\right)=-2+\ln(4) \\ 2\int_{0}^{1}\frac{x^2}{1-x^2}\mathrm{d}x-\int_{0}^{1}\frac{1}{1-x}\mathrm{d}x=-2+\ln(2) \end{align}
As far as computing $\Gamma'(1/2)$ goes, note that it equals $\Gamma(1/2)\psi(1/2)$, where $\Gamma(1/2)=\sqrt\pi$ is well-known and $\psi$ is the digamma function given by
\begin{align}\psi(1/2)&=-\gamma+\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/2}\right)\\&=-\gamma+2\sum_{n=1}^\infty\left(\frac1{2n}-\frac1{2n-1}\right)\\&=-\gamma-2\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\\&=-\gamma-2\ln(2)\\&=-\gamma-\ln(4)\end{align}
which is the proper result. Hence $\Gamma'(1/2)=-\sqrt\pi(\gamma+\ln(4))$.
As for what you've done, it is not valid to split a sum into two diverging sums, and hence this falls apart at
$$\sum_{n=1}^\infty\left(\frac2{1+2n}-\frac1n\right)\stackrel!=\sum_{n=1}^\infty\frac2{1+2n}-\sum_{n=1}^\infty\frac1n$$
which is simply false. This can be amended by simply not splitting the sum apart.