Section 4

Improper Integrals Between Infinite Limits

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Problem 2.30

Assume that the monotone function f(x)f(x) is defined for x≥0x \geq 0 and that the improper integral ∫0∞f(x) dx\int_{0}^{\infty } f(x) \, dx exists. Then we have

lim⁡h→+0h(f(h)+f(2h)+f(3h)+⋯ )=lim⁡h→+0h∑n=1∞f(nh)=∫0∞f(x) dx. \lim _{h \to +0} h \left(f(h) + f(2h) + f(3h) + \cdots \right) = \lim _{h \to +0} h \sum _{n=1}^{\infty } f(nh) = \int _{0}^{\infty } f(x) \, dx .
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Problem 2.31

The Γ\Gamma-function is defined for α>0\alpha > 0 (or ℜα>0\Re \alpha > 0) by the integral

Γ(α)=∫0∞e−xxα−1 dx. \Gamma (\alpha ) = \int _{0}^{\infty } e^{-x} x^{\alpha -1} \, dx .

Using Problem 89 of Part I prove that

Γ(α)=lim⁡n→∞nα−1n!α(α+1)⋯(α+n−1),α>0. \Gamma (\alpha ) = \lim _{n \to \infty } \frac{n^{\alpha -1} n!}{\alpha (\alpha +1) \cdots (\alpha +n-1)}, \qquad \alpha > 0 .
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Problem 2.32

As it is well known [cf. E. T. Whittaker and G. N. Watson, p. 246] Euler's constant CC can be written as an integral

C=∫0∞e−x(11−e−x−1x) dx. C = \int _{0}^{\infty } e^{-x} \left(\frac{1}{1 - e^{-x}} - \frac{1}{x}\right) \, dx .

Show that

C=lim⁡t→1−0[(1−t)(t1−t+t21−t2+t31−t3+⋯+tn1−tn+⋯ )−log⁡11−t]. C = \lim _{t \to 1-0} \left[(1-t) \left(\frac{t}{1-t} + \frac{t^{2}}{1-t^{2}} + \frac{t^{3}}{1-t^{3}} + \cdots + \frac{t^{n}}{1-t^{n}} + \cdots \right) - \log \frac{1}{1-t}\right] .
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Problem 2.33
lim⁡t→1−0(1−t)(t1+t+t21+t2+t31+t3+⋯+tn1+tn+⋯ )=log⁡2. \lim _{t \to 1-0} (1-t) \left(\frac{t}{1+t} + \frac{t^{2}}{1+t^{2}} + \frac{t^{3}}{1+t^{3}} + \cdots + \frac{t^{n}}{1+t^{n}} + \cdots \right) = \log 2 .
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Problem 2.34
lim⁡t→1−0(1−t)2(t1−t+2t21−t2+3t31−t3+⋯+ntn1−tn+⋯ )=π26. \lim _{t \to 1-0} (1-t)^{2} \left(\frac{t}{1-t} + 2 \frac{t^{2}}{1-t^{2}} + 3 \frac{t^{3}}{1-t^{3}} + \cdots + n \frac{t^{n}}{1-t^{n}} + \cdots \right) = \frac{\pi ^{2}}{6} .
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Problem 2.35

We have

lim⁡t→1−01−t (1+t+t4+t9+⋯+tn2+⋯ )=π2 \lim _{t \to 1-0} \sqrt{1-t} \, (1 + t + t^{4} + t^{9} + \cdots + t^{n^{2}} + \cdots ) = \frac{\sqrt{\pi }}{2}

and more generally for α>0\alpha > 0

lim⁡t→1−01−tα (1+t1α+t2α+t3α+⋯+tnα+⋯ )=1αΓ(1α). \lim _{t \to 1-0} \sqrt[\alpha ]{1-t} \, (1 + t^{1^{\alpha }} + t^{2^{\alpha }} + t^{3^{\alpha }} + \cdots + t^{n^{\alpha }} + \cdots ) = \frac{1}{\alpha } \Gamma \left(\frac{1}{\alpha }\right) .
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Problem 2.36

Compute

lim⁡t→+∞(1t+2tt2+12+2tt2+22+⋯+2tt2+n2+⋯ ). \lim _{t \to +\infty } \left(\frac{1}{t} + \frac{2t}{t^{2}+1^{2}} + \frac{2t}{t^{2}+2^{2}} + \cdots + \frac{2t}{t^{2}+n^{2}} + \cdots \right) .
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Problem 2.37

Let α>1\alpha > 1 and put g(t)=∏n=1∞(1+tnα)g(t) = \prod_{n=1}^{\infty } \left(1 + \dfrac {t}{n^{\alpha }}\right). Show that

lim⁡t→+∞log⁡g(t)t1α=πsin⁡πα. \lim _{t \to +\infty } \frac{\log g(t)}{t^{\frac{1}{\alpha }}} = \frac{\pi }{\sin \dfrac {\pi }{\alpha }} .
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Problem 2.38

Establish the equation

∫0∞log⁡(1−2x−2cos⁡2φ+x−4) dx=2πsin⁡φ,0≤φ≤π, \int _{0}^{\infty } \log (1 - 2x^{-2} \cos 2\varphi + x^{-4}) \, dx = 2\pi \sin \varphi , \qquad 0 \leq \varphi \leq \pi ,

with help of the following identity, valid for all complex tt:

sin⁡tt=eit−e−it2it=∏n=1∞(1−t2n2π2). \frac{\sin t}{t} = \frac{e^{it} - e^{-it}}{2it} = \prod _{n=1}^{\infty } \left(1 - \frac{t^{2}}{n^{2}\pi ^{2}}\right) .
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Problem 2.39

Compute the integral ∫0alog⁡x dx\int_{0}^{a} \log x \, dx as the limit of an infinite sum of rectangles corresponding to the points of division

a,aq,aq2,aq3,…,0<q<1. a, aq, aq^{2}, aq^{3}, \ldots , \qquad 0 < q < 1 .
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Problem 2.40

Let kk be a fixed positive number and nn an integer increasing to infinity. Then

∑v=0n(nv)k∼2knk(2πn)k−12 \sum _{v=0}^{n} \binom {n}{v}^{k} \sim \frac{2^{kn}}{\sqrt{k}} \left(\frac{2}{\pi n}\right)^{\frac{k-1}{2}}

(Problem 58). Observe the particular cases k=1k = 1, k=2k = 2.

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