The integral int_{frac{-1}{2}}^{frac{1}{2}} l | vedclass

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Question

(A) Let $I = \int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$. We can split the integral as: $I = \int_{\

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$-\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$. We can split the integral as: $I = \int_{\frac{-1}{2}}^{0} [x] dx + \int_{0}^{\frac{1}{2}} [x] dx + \int_{\frac{-1}{2}}^{\frac{1}{2}} \log_{e}\left(\frac{1+x}{1-x}\right) dx$. Let $g(x) = \log_{e}\left(\frac{1+x}{1-x}\right)$. Then $g(-x) = \log_{e}\left(\frac{1-x}{1+x}\right) = \log_{e}\left(\left(\frac{1+x}{1-x}\right)^{-1}\right) = -\log_{e}\left(\frac{1+x}{1-x}\right) = -g(x)$. Since $g(-x) = -g(x)$,$g(x)$ is an odd function. Therefore,$\int_{\frac{-1}{2}}^{\frac{1}{2}} g(x) dx = 0$. Now,for the greatest integer function $[x]$: In the interval $[-\frac{1}{2}, 0)$,$[x] = -1$. In the interval $[0, \frac{1}{2}]$,$[x] = 0$. Thus,$I = \int_{\frac{-1}{2}}^{0} (-1) dx + \int_{0}^{\frac{1}{2}} (0) dx + 0$. $I = [-x]_{\frac{-1}{2}}^{0} = -(0 - (-\frac{1}{2})) = -\frac{1}{2}$.

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$,where $[x]$ represents the greatest integer function,equals:

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