(1) ∫0∞e−xdx=[−e−x]0∞=0−(−1)=1\int_0^\infty e^{-x}dx = \left[-e^{-x}\right]_0^\infty = 0-(-1) = 1∫0∞e−xdx=[−e−x]0∞=0−(−1)=1 (2) ∫0∞xe−xdx=[x(−e−x)]0∞+∫0∞e−xdx=0+[−e−x]0∞=1\int_0^\infty xe^{-x}dx = \left[x(-e^{-x})\right]_0^\infty + \int_0^\infty e^{-x}dx = 0 + \left[-e^{-x}\right]_0^\infty = 1∫0∞xe−xdx=[x(−e−x)]0∞+∫0∞e−xdx=0+[−e−x]0∞=1 (3) ∫1edxxlogxlogx=t とおくと1xdx=dtx:1→et:0→1これから\int_1^e \frac{dx}{x\log x} \qquad \log x = t \ \text{とおくと} \quad \frac{1}{x}dx = dt \qquad x:1\to e \qquad t:0\to 1 \quad \text{これから}∫1exlogxdxlogx=t とおくとx1dx=dtx:1→et:0→1これから =∫011tdt=[logt]01=0−(−∞)= \int_0^1 \frac{1}{t}dt = \left[\log t\right]_0^1 = 0-(-\infty)=∫01t1dt=[logt]01=0−(−∞) よって発散する (4) ∫01logx dx=[xlogx]01−∫01x⋅1xdx=0−[x]01=−1\int_0^1 \log x\,dx = \left[x\log x\right]_0^1 - \int_0^1 x\cdot\frac{1}{x}dx = 0-\left[x\right]_0^1 = -1∫01logxdx=[xlogx]01−∫01x⋅x1dx=0−[x]01=−1