[jacks]

Evaluation of \( \displaystyle \int^\infty_0\frac{x+1}{x+2}\cdot \frac{x+3}{x+4}\cdot \frac{x+5}{x+6}\cdots\cdots dx \)

[Luis Fuentes]

Hola

Evaluation of \( \displaystyle \int^\infty_0\frac{x+1}{x+2}\cdot \frac{x+3}{x+4}\cdot \frac{x+5}{x+6}\cdots\cdots dx \)

I suppose that this means:

\( \displaystyle \int^\infty_0 f(x) dx \)

with:

\( f(x)=\displaystyle\lim_{n \to{+}\infty}{}\displaystyle\prod_{k=1}^n\dfrac{x+2k-1}{x+2k} \)

But:

\( (x+2k-1)(x+2k+1)=x^2+4k+(4k^2-1)<x^2+4k+4k^2=(x+2k)^2 \)

and from this:

\( \left(\displaystyle\prod_{k=1}^n\dfrac{x+2k-1}{x+2k}\right)^2=\dfrac{(x+1)(x+2n-1)}{(x+2n)^2}\displaystyle\prod_{k=1}^{n-1}\dfrac{(x+2k-1)(x+2k+1)}{(x+2k)^2}< \dfrac{(x+1)(x+2n-1)}{(x+2n)^2}\to 0 \)

So, in fact \( f(x)=0 \) for all \( x\geq 0 \) and the integral is zero.

Best regards.

[Juan Pablo Sancho]

Sabiendo la solución de antemano:
\( \displaystyle \prod_{k=1}^{+\infty} \dfrac{x+2k-1}{x+2k} = \prod_{k=1}^{+\infty} \dfrac{x+2k-1}{x+2k} = \prod_{k=1}^{+\infty} (1-\dfrac{1}{x+2k}) \leq \prod_{k=1}^{+\infty} e^{-\dfrac{1}{2k+x}}  \).

[jacks]

Thanks Administartor and Moderator.