The integrand is positive and continuous on the interval in question. Substitute

t = lo g (1 + x)

to obtain

\int_{1}^{\infty} f (x) d x = \int_{l o g 2}^{\infty} \frac{e ^{(1 - α) t}}{t ^{2}} d t .

α \geq 1

, the exponential factor is at most one and the comparison integral of

t^{- 2}

converges. If

α < 1

t^{2}

, so the integrand eventually exceeds one and the integral diverges. Thus

α \geq 1 .

The logarithmic square is what makes the boundary value

α = 1

integrable.

b

i

The function is measurable and

0 \leq cos^{2} (1/ x) \leq 1

. The interval has finite measure, so its integral is at most one. Hence it is integrable, despite its oscillations near zero.

x \to \infty

cos^{2} (1/ x) \to 1

. It is therefore at least

1/2

on some infinite tail. The integral of that lower bound is infinite, so it is not integrable on this interval.

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