数学高等数学《微积分》《微积分》 ∫0+∞e−ss5ds2+∫−∞+∞e−t22dt∫0+∞sint2dt⋅(∑n=0∞(−1)n2n+1∫0+∞sinxxdx+∑n=1∞arctan2n2limt→0+∫−20202020tcosxx2+t2dx)limn→∞(((∫01xn−11+xdx)n−12)⋅n2)=520\boxed{\frac{\displaystyle\frac{ \displaystyle\int_{0}^{+\infty} e^{-s} s^{5} \mathrm{d} s }{2}+\frac{ \displaystyle\int_{-\infty}^{+\infty} e^{-\frac{t^2}{2}} \mathrm{d} t }{ \displaystyle\int_{0}^{+\infty} \sin t^{2} \mathrm{d} t }\cdot\left(\frac{ \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1} }{ \displaystyle\int_{0}^{+\infty} \frac{\sin x}{x} \mathrm{d} x }+\frac{ \displaystyle\sum_{n=1}^{\infty} \arctan \frac{2}{n^2} }{ \displaystyle\lim _{t \to 0^+} \int_{-2020}^{2020} \frac{t \cos x}{x^2+t^2} \mathrm{d} x }\right)}{ \displaystyle\lim _{n \to \infty}\left(\left(\left(\int_{0}^{1} \frac{x^{n-1}}{1+x} \mathrm{d} x\right) n-\frac{1}{2}\right)\cdot\frac{n}{2}\right) }=520}n→∞lim(((∫011+xxn−1dx)n−21)⋅2n)2∫0+∞e−ss5ds+∫0+∞sint2dt∫−∞+∞e−2t2dt⋅∫0+∞xsinxdxn=0∑∞2n+1(−1)n+t→0+lim∫−20202020x2+t2tcosxdxn=1∑∞arctann22=520