Question

Evaluate log 1 upon x -1 (dx)​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle\int\limits^1_0 {log\bigg(\dfrac{1}{x} - 1 \bigg) } \, dx$

As we know that,

First we simplify the equation, we get.

$\sf \implies \displaystyle\int\limits^1_0 log \bigg(\dfrac{1 - x}{x} \bigg)dx \ \ \ ---(1)$

As we know that,

Formula of :

$\sf \implies \displaystyle\int\limits^a_0 f(x)dx = \displaystyle\int\limits^a_0 f(a - x)dx$

Using this formula in equation, we get.

$\sf \implies \displaystyle\int\limits^1_0 log \bigg(\dfrac{1 - (1 - x)}{1 - x} \bigg)dx.$

$\sf \implies \displaystyle\int\limits^1_0 log \bigg(\dfrac{1 - 1 + x}{1 - x} \bigg)dx$

$\sf \implies \displaystyle\int\limits^1_0 log\bigg(\dfrac{x}{1 - x} \bigg)dx \ \ \ \ -----(2)$

From equation (1) & (2), we get.

Adding equation (1) & (2), we get.

$\sf \implies I + I = \displaystyle\int\limits^1_0 log \bigg(\dfrac{1 - x}{x} \bigg)dx \ \ + \displaystyle\int\limits^1_0 log \bigg(\dfrac{x}{1 - x} \bigg)dx$

$\sf \implies 2I = \displaystyle\int\limits^1_0 log \bigg[ \dfrac{1 - x}{x} \ \times \ \dfrac{x}{1 - x} \bigg]dx$

$\sf \implies 2I = \displaystyle\int\limits^1_0 log (1)dx$

$\sf \implies 2I = 0.$

$\sf \implies I = 0.$

                                                                                                                       

MORE INFORMATION.

Summation of series by integration.

For finding sum of an infinite series with the help of definite integration, following formula is used.

$\sf \implies \displaystyle \lim_{n \to \infty} \sum \limits_{r = 0}^{n - 1} f\bigg(\frac{r}{n}\bigg)\bigg(\frac{1}{n} \bigg) = \int\limits^1_0 {f(x)} \, dx$

Working rules.

(1) = Express the given series in the form of = $\sf \implies \displaystyle \lim_{n \to \infty} \sum \limits_{r = 0}^{n - 1} f\bigg(\frac{r}{n}\bigg)\bigg(\frac{1}{n} \bigg)$

(2) = Replace (r/n) by x, 1/n by dx and ∑ by ∫, we get the integral ∫f(x)dx in place of above series.

(3) = The lower limit of this integrals = $\sf \displaystyle \lim_{n \to \infty} \bigg(\frac{r}{n} \bigg)_{r = 0}$  r is least value in this case r = 0.

(4) = Upper limit = $\sf \displaystyle \lim_{n \to \infty} \bigg(\frac{r}{n}\bigg)_{r = n - 1}$  r is greatest value in this case r = n - 1.