SOLUTIONS

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6. Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

Prove

∑_{n≤x} \frac{1}{3n − 2} = \frac{1}{3} log(3x − 2) + \frac{1}{6} log 3 + \frac{π}{6\sqrt{3}} + \frac{γ}{3} + O(\frac{1}{x}) ,

where log is the natural log and γ is Euler’s constant.

Solution by the proposers.

We start with the following lemma. \underline{Lemma}.

∫ ∞_{\frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du = 1 − γ − \frac{π}{2\sqrt{3}} − \frac{1}{2} log 3 .

\underline{Proof}.

∫ ∞_{\frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du = ∫ 1_{\frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du + ∑_{n=1}^{∞} ∫_{n}^{n+1} \frac{u − [u + \frac{2}{3}]}{u^{2}} du

= ∫ 1_{\frac{1}{3}} \frac{u − 1}{u^{2}} du + ∑_{n=1}^{∞}(∫_{n}^{n+ \frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du + ∫_{n+ \frac{1}{3}}^{n+1} \frac{u − [u + \frac{2}{3}]}{u^{2}} du)

= log u + \frac{1}{u}|1_{\frac{1}{3}} + ∑_{n=1}^{∞}(∫_{n}^{n+ \frac{1}{3}} \frac{u − [u]}{u^{2}} du + ∫_{n+ \frac{1}{3}}^{n+1} \frac{u − [u] − 1}{u^{2}} du)

= log 3 − 2 + ∑_{n=1}^{∞}(∫_{n}^{n+1} \frac{u − [u]}{u^{2}} du − ∫_{n+ \frac{1}{3}}^{n+1} \frac{du}{u^{2}})

= log 3 − 2 + ∫_{1}^{∞} \frac{u − [u]}{u^{2}} du + ∑_{n=1}^{∞} \frac{1}{u}|_{n+ \frac{1}{3}}^{n+1}

= log 3 − 2 + ∫_{1}^{∞} \frac{u − [u]}{u^{2}} du + ∑_{n=1}^{∞}(\frac{1}{n + 1} − \frac{1}{n + \frac{1}{3}} )

= log 3 − 2 + (1 − γ) + (2 − \frac{π}{2\sqrt{3}} − \frac{3}{2} log 3)

= 1 − γ − \frac{π}{2\sqrt{3}} − \frac{1}{2} log 3 .

The next to last equality follows from

∫_{1}^{∞} \frac{u − [u]}{u^{2}} du = 1 − γ [1]

and

∑_{n=1}^{∞}(\frac{1}{n + 1} − \frac{1}{n + \frac{1}{3}} ) = 2 − \frac{π}{2\sqrt{3}} − \frac{3}{2} log 3 [2] .

To prove the result, we have by Euler’s summation formula [1] that

∑_{n≤x} \frac{1}{3n − 2} = 1 + ∫_{1}^{x} \frac{dt}{3t − 2} − 3 ∫_{1}^{x} \frac{t − [t]}{(3t − 2)^{2}} dt + \frac{[x] − x}{3x − 2}

= \frac{1}{3} log(3x − 2) + 1 − 3 ∫_{1}^{x} \frac{t − [t]}{(3t − 2)^{2}} dt + O(\frac{1}{x})

= \frac{1}{3} log(3x − 2) + 1 − 3 ∫_{1}^{∞} \frac{t − [t]}{(3t − 2)^{2}} dt + 3 ∫_{x}^{∞} \frac{t − [t]}{(3t − 2)^{2}} dt + O(\frac{1}{x})

= \frac{1}{3} log(3x − 2) + 1 − 3 ∫_{1}^{∞} \frac{t − [t]}{(3t − 2)^{2}} dt + O(\frac{1}{x}) .

Next,

∫_{1}^{∞} \frac{t − [t]}{(3t − 2)^{2}} dt = \frac{1}{9} ∫_{1}^{∞} \frac{t − [t]}{(t − \frac{2}{3})^{2}} dt

= \frac{1}{9} ∫ ∞_{\frac{1}{3}} \frac{u + \frac{2}{3} − [u + \frac{2}{3}]}{u^{2}} du

= \frac{1}{9} ∫ ∞_{\frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du + \frac{2}{27} ∫ ∞_{\frac{1}{3}} \frac{du}{u^{2}}

= \frac{1}{9} ∫ ∞_{\frac{1}{3}} \frac{u − [u + \frac{2}{3}]}{u^{2}} du + \frac{2}{9} .

Thus by the lemma,

\begin{aligned}∫_{1}^{∞} \frac{t − [t]}{(3t − 2)^{2}} dt & = \frac{1}{9}(1 − γ − \frac{π}{2\sqrt{3}} − \frac{1}{2} log 3) + \frac{2}{9} \\ & = \frac{1}{3} − \frac{γ}{9} − \frac{π}{18\sqrt{3}} − \frac{1}{18} log 3 , \end{aligned}

so

∑_{n≤x} \frac{1}{3n − 2} = \frac{1}{3} log(3x − 2)

+ 1 − 3(\frac{1}{3} − \frac{γ}{9} − \frac{π}{18\sqrt{3}} − \frac{1}{18} log 3) + O(\frac{1}{x})

= \frac{1}{3} log(3x − 2) + \frac{1}{6} log 3 + \frac{π}{6\sqrt{3}} + \frac{γ}{3} + O(\frac{1}{x}) .

\underline{References}

1. T. M. Apostol, Introduction to Analytic Number Theory, Springer–Verlag, 1984, pp. 56,54.

2. D. E. Knuth, The Art of Computer Programming: Fundamental Algorithms, Vol. 1, Second Edition, Addison–Wesley, p. 94.

9. Proposed by Robert E. Shafer, Berkeley, California.

Let p be a prime. Prove that

(p−1)! ≡ (p+1)(1^{p−1} +2^{p−1} +3^{p−1} +⋯+(p−1)^{p−1}) (mod p^{2}) .

Solution by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

The problem as stated above is incorrect. Letting p = 2 yields

a counterexample. In the remainder of our solution we shall assume

that p is an odd prime number.

We shall use the following known result. If p is an odd prime number, then

(∗) 1^{p−1} + 2^{p−1} + ⋯ + (p − 1)^{p−1} ≡ p + (p − 1)! (mod p^{2}) .

[For a proof of (*), see exercise 5 and its solution on p. 211 of Sierpinski, \underline{Elementary Theory of Numbers}, Hafner Publishing Company, New York, 1964.] We may now construct the following collection of equivalent congruences.

(p − 1)! ≡ (p + 1)(1^{p−1} + 2^{p−1} + ⋯ + (p − 1)^{p−1}) (mod p^{2})
(p − 1)! ≡ (p + 1)(p + (p − 1)!) (mod p^{2}) [by (∗)]
(p − 1)! ≡ p^{2} + p(p − 1)! + p + (p − 1)! (mod p^{2})
p(p − 1)! ≡ −p (mod p^{2})
(p − 1)! ≡ −1 (mod p) .

But (p − 1)! ≡ −1 (mod p) follows from Wilson’s Theorem so our solution is complete.

Also solved by the proposer.

12. Proposed by Robert E. Shafer, Berkeley, California.

Evaluate

∫_{0}^{1} e^{1/log x} dx ,

where log denotes the natural log.

Solution by Mark Ashbaugh, University of Missouri, Columbia, Missouri.

The integral evaluates to 2K_{1}(2) ≈ .2797318 where K_{1} denotes one of the two standard modified Bessel functions of order 1. It would be surprising though perhaps not impossible for there to be a more elementary expression for the value of this integral.

Let I denote the given integral. Then under the substitution t = −1/ log x it becomes

\begin{aligned}I & = ∫_{0}^{∞} t^{−2}e^{−(t+1/t)} dt \\ & = ∫_{0}^{1} t^{−}2e^{−(t+1/t)} dt + ∫_{1}^{∞} t^{−2}e^{−(t_{1}/t)} dt . \end{aligned}

Now change variables to s = 1/t in the first of these integrals to arrive at

\begin{aligned}I & = ∫_{1}^{∞} e^{−(s+1/s)} ds + ∫_{1}^{∞} t^{−2}e^{−(t+1/t)} dt \\ & = ∫_{1}^{∞}(1 + t^{−2})e^{−(t+1/t)} dt . \end{aligned}

Finally, put t = e^{u} in this integral, obtaining

I = 2 ∫_{0}^{∞} cosh ue^{−2 cosh u} du .

Since one of the standard integral representations for the modified Bessel function K_{ν}(z) is

K_{ν}(z) = ∫_{0}^{∞} e^{−z cosh t} cosh(νt) dt

[Abramowitz and Stegun, p. 376, eq. 9.6.24] we can identify the value of our integral as

\begin{aligned}I & = 2K_{1}(2) \\ & = 2[γI_{1}(2) + \frac{1}{2} − ∑_{n=0}^{∞} \frac{H_{n} + H_{n+1}}{n!(n + 1)!} ] \end{aligned}

where γ ≈ .5772157 is Euler’s constant,

H_{n} ≡ ∑_{m=1}^{n} \frac{1}{m}

with the convention that H_{0} = 0, and I_{1} represents the other standard modified Bessel function of order 1 and thus, in particular,

I_{1}(2) = ∑_{n=0}^{∞} \frac{1}{n!(n + 1)!} .

In an analogous fashion one can show that the function I(a) defined by the integral

I(a) = ∫_{0}^{1} e^{a/ log x} dx for a > 0

is given by

I(a) = 2\sqrt{a}K_{1}(2\sqrt{a}) .

The sequence of substitutions to be used is now t = −\sqrt{a}/ log x , s = 1/t , and t = e^{u} .

Also solved by the proposer.