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5.10 Integration over unbounded regions

We can integrate ff over regions RR more general than rectangles by chopping up RR into sets of small diameter and approximating by rectangles. It is also possible to allow RR to be an unbounded region, provided that ff satisfies suitable conditions.

Example.

To show that

∫01∫0∞x⁢sin⁡x⁢y⁢e-y⁢d⁢y⁢d⁢x=1-π4.\int_{0}^{1}\!\!\!\int_{0}^{\infty}x\sin xy\,e^{-y}\,dy\,dx=1-{{\pi}\over{4}}.

Solution. We note that the inner integral ∫0∞x⁢sin⁡x⁢y⁢e-y⁢d⁢y\int_{0}^{\infty}x\sin xy\,e^{-y}\,dy can be immediately described using the Laplace transform of sin⁡a⁢x\sin ax (or rather, sin⁡a⁢y\sin ay): it is x2x2+1\frac{x^{2}}{x^{2}+1}. (Alternatively, we can integrate by parts as in 1.47-8.)

Thus the double integral is

∫01x2x2+1⁢d⁢x=∫01(1-1x2+1)⁢d⁢x=[x-tan-1⁡x]01= 1-π4.\int_{0}^{1}\frac{x^{2}}{x^{2}+1}\,dx=\,{\int_{0}^{1}\left(1-\frac{1}{x^{2}+1}% \right)\,dx=}\,{\left[x-\tan^{-1}x\right]_{0}^{1}=}\,{1-\frac{\pi}{4}.}