Calcular

Question

Estoy tratando de calcular la siguiente integral:

PS

Intenté el cambio habitual t = 1 / x pero no pude avanzar más.

¡Gracias!

Answer 1

$$\int_{1}^{\infty}\frac{\lceil x\rceil}{x^3}\,dx =\sum_{n\geq 1}\int_{n}^{n+1}\frac{n+1}{x^3}\,dx=\sum_{n\geq 1}\frac{2n+1}{2n^2(n+1)}=\color{red}{\frac{1}{2}+\frac{\pi^2}{12}}$ $ se mantiene por descomposición parcial de la fracción.

Answer 2

Si $\frac 1{n} \le x < \frac 1{n-1}$ entonces $\lceil \frac 1x \rceil = n$ . y $\int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx= \int_{\frac 1{n}}^{\frac 1{n-1}} xn dx = n\frac {x^2}2|_{\frac 1n}^{\frac 1{n-1}}= \frac n2(\frac 1{(n-1})^2 -\frac 1{n^2}=\frac n2(\frac {2n-1}{n^2(n-1)^2})=\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}$

Entonces $\int_{0}^{1} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty \int_{\frac 1{n}}^{\frac 1{n-1}} x\lceil \frac 1x \rceil dx=\sum\limits_{n=2}^\infty[\frac 1{(n-1)^2} - \frac {1}{2n(n-1)^2}]=\sum\limits_{n=1}^\infty[\frac 1{n^2} - \frac {1}{2(n+1)n^2}]$

Answer 3

$$ \begin{aligned} \int_0^1x\cdot\left\lceil \frac{1}{x}\right\rceil\; dx &= \sum_{k\ge 1} \int_{1/(k+1)}^{1/k}x\cdot \underbrace{\left\lceil \frac{1}{x}\right\rceil}_{\in (k,k+1)\text{ a.e}} \;dx \\ &= \sum_{k\ge 1} \int_{1/(k+1)}^{1/k}x\cdot (k+1)\;dx \\ &= \sum_{k\ge 1} (k+1) \left[\ \frac 12 x^2\ \right]_{1/(k+1)}^{1/k} \\ &= \frac 12 \sum_{k\ge 1} (k+1) \left[\ \frac 1{k^2}-\frac 1{(k+1)^2}\ \right]_{1/(k+1)}^{1/k} \\ &= \frac 12 \sum_{k\ge 1} \frac {2k+1}{k^2(k+1)} \\ &= \frac 12 \sum_{k\ge 1} \frac k{k^2(k+1)} + \frac 12 \sum_{k\ge 1} \frac {k+1}{k^2(k+1)} \\ &= \frac 12 \sum_{k\ge 1} \left[ \frac 1k-\frac 1{k+1}\right] + \frac 12 \sum_{k\ge 1} \frac 1{k^2} \\ &= \frac 12\cdot 1+\frac 12\cdot\frac{\pi^2}6\ . \end {alineado} $$ Verificación de computadora, aquí PARI / GP:

 ? intnum( x=1.e-12, 1, x*ceil(1/x), 8 )
%12 = 1.3224146303918600341384604269228003273
? 1/2 + Pi^2/12
%13 = 1.3224670334241132182362075833230125946