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1.42 Integrals over infinite ranges

1.42.1 $\int_{0}^{\infty}e^{-x}\,dx=1.$

Solution. We have $\int_{0}^{R}e^{-x}\,dx=$ $[-e^{-x}]_{0}^{R}=1-e^{-R}$ $\rightarrow 1$ as $R\rightarrow\infty$ .

1.42.2 The integral $\int_{1}^{\infty}x^{-s}\,dx=1/(s-1)$ converges for $s>1$ ; whereas for $s\leq 1$ the integral diverges.

Solution. We have $\int_{1}^{R}x^{-s}\,dx=$ $[-\frac{x^{1-s}}{s-1}]_{1}^{R}=\frac{1-\frac{1}{R^{s-1}}}{s-1}$ .

For $s>1$ : $\frac{1}{R^{s-1}}\rightarrow 0$ as $R\rightarrow\infty$ , hence the integral converges to $\frac{1}{s-1}$ .

If $s<1$ , then $\frac{1}{R^{s-1}}\rightarrow\infty$ as $R\rightarrow\infty$ , so the integral diverges.

Finally: $\int_{1}^{R}x^{-1}\,dx=$ $[|\log x|]_{1}^{R}$ $=\log R-\log 1\rightarrow\infty$ as $R\rightarrow\infty$ .