I've tried many ways, but it seems that it didn't work: $$ \int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)} = \int_0^1\frac{x^{n-1}}{(x+1)(2x+1)\cdots(nx+1)}\mathrm dx = \cdots $$ Any help would be appreciated!

  • $\begingroup$ I think your transformation is wrong. The denominator would have terms like $1+k x$, not $x+k$ after transforming to $[0,1]$. $\endgroup$
    – Ron Gordon
    Dec 13 '14 at 12:18
  • $\begingroup$ @RonGordon Thanks for your reminder, I've updated the question yet. $\endgroup$
    – Shine Mic
    Dec 13 '14 at 12:24
  • $\begingroup$ Have you tried any methods from contour integration? $\endgroup$
    – Myself
    Dec 13 '14 at 12:26
  • $\begingroup$ Partial fraction decomposition leads to a sum with $n$ terms. Would that be acceptable, or do you need something without such a sum? $\endgroup$ Dec 13 '14 at 12:32
  • $\begingroup$ @DanielFischer Doesn't the integral have a closed-form expression which is related to $n$? If not, the sum with $n$ terms is acceptable of course. $\endgroup$
    – Shine Mic
    Dec 13 '14 at 12:37

We have the partial fraction decomposition $$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac{a_0}{x}+\frac{a_1}{x+1}+\frac{a_2}{x+2}+\ldots+\frac{a_n}{x+n}$$

Now as $x\to 0$, the part with $a_0$ dominates, and we have $$a_0=\lim_{x\to0}\frac{x}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}$$ Similarily, as $x\to -1$, the part with $a_1$ dominates, and we have $$a_1=\lim_{x\to-1}\frac{x+1}{x(x+1)(x+2)\cdots(x+n)}=-\frac1{(n-1)!}$$ and if you were to continue this idea, you'd find that $$a_k=\lim_{x\to -k}\frac{x+k}{x(x+1)(x+2)\cdots(x+n)}= \frac{(-1)^k}{k!(n-k)!}=\frac{(-1)^k {n\choose k}}{n!}$$ so that $$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)} $$ and so $$\int_1^{\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)} =\frac{1}{n!}\int_1^{\infty}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)}\\ =\frac1{n!}\left(\sum_{k=0}^n(-1)^k {n\choose k}\ln(x+k)\right)\left. \right|_1^\infty$$ For $x=\infty$, consider the binomial expansion of $(1-1)^k$. It implies that the sum of the positive binomial terms is equal to the sum of the negative binomial terms, and so if we combine all the logs into one, we get the logarithm of a polynomial divided by a polynomial of the same degree, so the limit to infinity is hence $\ln(1) = 0$. So then the integral is thus equal to the negative of the antiderivative above at $x=1$, so $$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n(-1)^{k+1} {n\choose k}\ln(k+1)\\=\frac1{n!}\sum_{k=2}^{n+1}(-1)^{k} {n\choose k-1}\ln(k)$$


If you sub $x=u+1$ then you get

$$\int_0^{\infty} \frac{dx}{(x+1)(x+2)\cdots(x+n+1)} $$

Try evaluating via residue theorem. Consider the contour integral

$$\oint_C dz\frac{\log{z}}{(z+1)(z+2)\cdots(z+n+1)} $$

where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. Take the limits $R\to \infty$ and $\epsilon \to 0$ and get that the contour integral is

$$-i 2 \pi \int_0^{\infty} \frac{dx}{(x+1)(x+2)\cdots(x+n+1)} $$

This is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand. In this case, we obviously have poles at $-1 \ldots-n$. Keep in mind the branch cut we took using $C$ so the negative is actually $e^{i \pi}$. The integral is then, taking respective real parts,

$$\begin{align} \int_0^{\infty} \frac{dx}{(x+1)(x+2)\cdots(x+n+1)} &= -\sum_{k=1}^{n+1} \frac{\log{k}}{\displaystyle \prod_{j=1 \cap j \ne k}^{n+1} (-j+k)}\\ &= \sum_{k=2}^{n+1} (-1)^{k} \frac{\log{k}}{(k-1)!(n+1-k)! } \\ &= \frac1{(n+1)!} \sum_{k=2}^{n+1} (-1)^{k} \binom{n+1}{k}k \log{k}\end{align}$$


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