求二重级数和

青青子衿

\[\huge
\begin{align*}
\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{(-1)^{m+n}}{m+n}&=\ln2-\frac{1}{2}\\
\\
\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\frac{(-1)^{m+n}\ln(m+n)}{m+n}&=\frac{1}{2}\ln\frac{\pi}{2}+\frac{1}{2}\ln^2{2}-\gamma\ln{2}
\end{align*}
\]

\begin{align*}
\int_0^b\int_0^a\frac{x^2+y^2}{x+y}{\rm\,d}x{\rm\,d}y=
\end{align*}

\begin{align*}
\Huge\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{\cos\left(ax\right)\cos\left(by\right)}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}{\kern 2pt}{\rm\,d}x{\rm\,d}y
\end{align*}

tommywong

$\displaystyle f(x)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^{m+n}}{m+n},f'(x)=\sum_{n=1}^\infty \sum_{m=1}^\infty x^{m+n-1}=\sum_{n=1}^\infty \frac{x^n}{1-x}=\frac{x}{(1-x)^2}$

$\displaystyle \int \frac{x}{(1-x)^2}dx=\int\frac{u-1}{u^2}du=lnu+\frac{1}{u}+C=ln(1-x)+\frac{1}{1-x}+C$

$\displaystyle f(x)=ln(1-x)+\frac{1}{1-x}-ln(1-0)-\frac{1}{1-0}=ln(1-x)+\frac{x}{1-x}$