improper integral

**ch2kb0x** · April 23rd 2008, 08:34 AM

Urgent Help!

integral from 1 to infinity (8ln(x)/x^2))dx

and would the series of that be convergent?

**Plato** · April 23rd 2008, 08:45 AM

Use the fact that $\ln (x) \le 2\sqrt x$ and simple comparsion.

**Soroban** · April 23rd 2008, 09:29 AM

Hello, ch2kb0x!

$\int^{\infty}_1\frac{8\ln x}{x^2}\,dx$

Integrate by parts . . .

. . $\begin{array}{ccccccc}u &=& 8\ln x & & dv &=& x^{-2}dx \\ du &=& \frac{8\,dx}{x} & & v & = &\text{-}\frac{1}{x} \end{array}$

And we have: . $-\frac{8}{x}\ln x + \int\frac{8}{x^2}\,dx \;\;=\;\;-\frac{8}{x}\ln x - \frac{8}{x} \;\;=\;\;-\frac{8}{x}(\ln x + 1)\,\bigg]^{\infty}_1$

Then: . $\lim_{b\to\infty}\bigg[-\frac{8}{x}(\ln x + 1)\,\bigg]^b_1 \;\;=\;\; \lim_{b\to\infty}\left[-\frac{8}{b}(\ln b + 1) + \frac{8}{1}(\ln 1 + 1)\right]$

. . . $= \;\lim_{b\to\infty}\bigg[-8\,\frac{\ln b + 1}{b} + 8\bigg] \quad\Rightarrow\quad -8\!\cdot\!\frac{\infty}{\infty} + 8$

Apply l'Hopital: . $\lim_{b\to\infty}\bigg[-8\,\frac{\frac{1}{b}}{1} + 8\bigg]\;\;=\;\;\lim_{b\to\infty}\left(-\frac{8}{b} + 8\right) \;\;=\;\;0 + 8 \;\;=\;\;\boxed{8}$

Thread: improper integral

LinkBack

Thread Tools

Display