18.01 PROBLEM SET 8 – SOLUTIONS

2. First compute the partial fraction decomposition of \frac{1+2t}{t^{2}(t+1)^{2}} .

\frac{1 + 2t}{t^{2}(t + 1)^{2}} = \frac{1}{t^{2}} − \frac{1}{(t + 1)^{2}} .

Next compute

∫_{x}^{z} \frac{1 + 2t}{t^{2}(t + 1)^{2}} dt = ∫_{x}^{z} \frac{dt}{t^{2}} − ∫_{x}^{z} \frac{dt}{(t + 1)^{2}} =

= −\frac{1}{t}|_{x}^{z} − (−\frac{1}{t + 1}) |_{x}^{z} = −\frac{1}{z} + \frac{1}{x} + \frac{1}{z + 1} − \frac{1}{x + 1}.

Therefore

∫_{x}^{∞} \frac{1 + 2t}{t^{2}(t + 1)^{2}} dt = lim_{z→∞} (−\frac{1}{z} + \frac{1}{x} + \frac{1}{z + 1} − \frac{1}{x + 1}) = \frac{1}{x} − \frac{1}{x + 1}.

Then

I = ∫_{0}^{∞} \frac{1}{x} − \frac{1}{x + 1}dx = lim_{v→0} ∫_{v}^{1} \frac{1}{x} − \frac{1}{x + 1}dx + lim_{w→∞} ∫_{1}^{w} \frac{1}{x} − \frac{1}{x + 1}dx =

= lim_{v→0} (ln x − ln(x + 1)) |_{v}^{1} + lim_{w→∞} (ln x − ln(x + 1)) |_{1}^{w} = = lim_{v→0}(ln 1 − ln v − ln 2 + ln(v + 1)) + lim_{w→∞}(ln w − ln 1 − ln(w + 1) + ln 2) = = lim_{v→0}(− ln v + ln(v + 1)) + lim_{w→∞}(ln w − ln(w + 1)) =

= lim_{v→0} ln (\frac{v + 1}{v}) + lim_{w→∞} ln (\frac{w}{w + 1}) = lim_{u→1} ln (u) + lim_{w→∞} ln (1 − \frac{1}{w + 1}) = ∞.

(15 points)

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