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Let $n\in\mathbb{N}$ and define $$f_n(x)=\prod_{k=0}^n(x+k)^{(-1)^k \binom{n}{k}}$$

Initially, I was interested in a variant of the multiple-angle formula for tangent, which involves (alternating) binomial-weighted powers of tangent in the numerator and denominator; ex, with $T=\tan(x),$ $\tan(4x) = (4T-4T^3)/(1-6T^2+T^4)$. The $f_n$ share a similar structure, but are purely multiplicative.

I am interested in $\lim_{n\to\infty} \int_0^1f_n(x)\,dx$. I am aware of several properties of the integrand:

  • $0\le f_n(x)\le 1$ for all $x\in [0,1]$.
  • Further, for all $x\in [0,1]$, $f_n(x)\le f_{n+1}(x)$. Thus, by the monotone convergence theorem, we conclude that $f_n$ converges pointwise to some $f$, and then the dominated convergence theorem guarantees $$\lim_{n\to\infty} \int_0^1f_n(x)= \int_0^1 \lim_{n\to\infty} f_n(x) $$
  • If we take natural log, we have $$(-1)^n \Delta^n \log(x) = \log(f_n(x)),$$where $\Delta$ is forward difference; see here.
  • The partial fractions decomposition of $f_n$ features factors of $(x+(2k+1))$ and powers of $2$ in the denominator, but the numerators and overall pattern are not apparent to me. For instance, $$f_5(x)=1+\frac{516045}{1048576(x+1)}-\frac{74205}{262144(x+1)^2}-\frac{39555}{65536(x+1)^3}-\frac{5265}{16384(x+1)^4}-\frac{243}{4096(x+1)^5}-\frac{1725}{8192(x+3)}-\frac{20715}{16384(x+3)^2}+\frac{915}{1024(x+3)^3}+\frac{2355}{2048(x+3)^4}-\frac{153}{128(x+3)^5}-\frac{195}{512(x+3)^6}+\frac{45}{64(x+3)^7}-\frac{15}{256(x+3)^8}-\frac{5}{32(x+3)^9}+\frac{3}{64(x+3)^{10}}-\frac{295245}{1048576(x+5)}$$
  • A related integral: $$n! \int\prod_{k=0}^{n} (x+k)^{-1} \,dx = \log(f_n(x))$$
  • Not exactly related, but similar:

Mathematical gives $\{0.306853,0.5,0.589549,0.641414,0.675638,0.700157,0.718736,0.733397,0.745323,0.755259\}$ as $\int_0^1 f_n(x)\,dx$, $1\le n \le 10$, but was unable to produce further values. I wonder if there is a clever way to evaluate the integral in question using the difference relationship mentioned above.

Integrand
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1 Answers1

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The limit equals $1$ since $\lim\limits_{n\to\infty}f_n(x)=1$ for $x>0$. To see the latter, use Frullani integral $$ \log z=\int_0^\infty\frac{e^{-t}-e^{-zt}}{t}\,dt.\qquad(z>0) $$ We obtain (for $n>0$) $$ \log f_n(x)=\sum_{k=0}^n(-1)^k\binom nk\log(x+k)=-\int_0^\infty\frac{e^{-xt}}{t}(1-e^{-t})^n\,dt, $$ which tends to $0$ monotonically as $n\to\infty$.

metamorphy
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  • Thanks for the answer; I hadn't thought about Frullani integrals, though they are quite natural. Convergence in $n$ is somewhat slow, which is why I thought the limit might be lower than $1$, but your last integrals makes it clear. – Integrand May 30 '25 at 15:03