EVALUATION OF A FAMILY OF IMPROPER INTEGRALS

Russell Euler Northwest Missouri State University

It is easy to see that the improper integral

(1)  I(q) = ∫_{1}^{∞} \frac{dx}{x(x^{q−1} + ⋯ + x + 1)}

converges if q ≥ 2. The purpose of this paper is to evaluate (1) when q is an integer.

It has been shown in [1], page 37, that

(2)  ψ(a) − ψ(a − b) = \frac{Γ(a)}{Γ(b)} ∑_{n=1}^{∞} \frac{Γ(b + n)}{nΓ(a + n)}

for Re(a) > Re(b) ≥ 0. Equation (2) will be of particular interest when the parameters are specialized by letting a = 1 and b = \frac{1}{q} for q = 2, 3, 4, ... . For then, (2) becomes

\frac{Γ(1)}{Γ (\frac{1}{q})} ∑_{n=1}^{∞} \frac{Γ(\frac{1}{q} + n)}{nΓ(1 + n)} = ψ(1) − ψ (1 − \frac{1}{q}) ,

and so

(3)  ∑_{n=1}^{∞} \frac{(\frac{1}{q})n}{n! n} = −γ − ψ (1 − \frac{1}{q}) ,

where γ is the Euler-Mascheroni constant and

(r)_{n} = \frac{Γ(r + n)}{Γ(r)} = r(r + 1) ⋯ (r + n − 1) .

Now, from page 13 of [2], it is known that (4)

ψ (1 − \frac{1}{q}) = − γ − ln q − \frac{π}{2} cot\frac{(q − 1)π}{q}

+ \begin{matrix}[ & \frac{q}{2} & ] \\ ∑ \\ n=1 \end{matrix} cos\frac{2n(q − 1)π}{q} ln (2 − 2cos\frac{2nπ}{q} ) ,

where the prime attached to the summation index indicates that if q is even then the last term in the sum is taken at half its weight. Substituting (4) into (3) and simplifying gives

(5)

∑_{n=1}^{∞} \frac{1(1 + q)(1 + 2q) ⋯ (1 + (n − 1)q)}{n! q^{n}n} = ln q

+ \frac{π}{2} cot\frac{(q − 1)π}{q} − 2 \begin{matrix}[ & \frac{q}{2} & ] \\ ∑ \\ n=1 \end{matrix} cos\frac{2n(q − 1)π}{q} ln (2|sin\frac{nπ}{q} |) .

Since

(1 − u)^{− \frac{1}{q}} = ∑_{n=0}^{∞} \binom{−\frac{1}{q}}{n}(−u)^{n} for − 1 ≤ u < 1 ,

∫_{0}^{1} \frac{(1 − u)^{− \frac{1}{q}} − 1}{u}du = ∑_{n=1}^{∞} \binom{−\frac{1}{q}}{n}(−1)^{n} ∫_{0}^{1} u^{n−1}du

= ∑_{n=1}^{∞} \frac{(−1)^{n} (−\frac{1}{q}) (−\frac{1}{q} − 1) (−\frac{1}{q} − 2) ⋯ (−\frac{1}{q} − n + 1)}{n! n}

= ∑_{n=1}^{∞} \frac{1(1 + q)(1 + 2q) ⋯ (1 + (n − 1)q)}{n! q^{n}n} .

In (6), let

x = (1 − u)^{− \frac{1}{q}} .

Then

u = 1 − x^{−q} , du = q x^{−q−1} dx ,

and so (7)

q ∫_{1}^{∞} \frac{(x − 1)dx}{(1 − x^{−q}) x^{q+1}} = ∑_{n=1}^{∞} \frac{1 (1 + q) ⋯ (1 + (n − 1)q)}{n! q^{n} n}.

Simplifying (7) and using (5) yields

I(q) = \frac{1}{q}(ln q + \frac{π}{2} cot\frac{(q − 1)π}{q}

− 2 \begin{matrix}[ & \frac{q}{2} & ] \\ ∑ \\ n=1 \end{matrix} cos\frac{2n(q − 1)π}{q} ln (2|sin\frac{nπ}{q} |)) .

\underline{References}

1. A. McBride and G. Roach, Fractional Calculus, Research Notes in Mathematics, Pitman Advanced Publishing Program, 1985.

2. Y. Luke, The Special Functions and Their Approximations, Vol. 1, Academic Press, 1969.