The integral integral _0^1 frac17^[(1 / x)] dx, where [.] denotes the greatest integer function is equal to

integral _0^1 frac17^[(1 / x)] dx=-integral _1^0 frac17^[(1 / x)] dx \ = (-1)[integral _1^1 / 2 (1 / 7) dx+integral _1 / 2^1 / 3 (1 / 7^2) dx+integral _1 / 3^1 / 4 (1 / 7^3) dx+ ... ... infinity] \ = ((1 / 7)+(1 / 2.7^2)+(1 / 3.7^3)+ ... infinity)-((1 / 7.2)+(1 / 7^2 .3)+(1 / 7^2 .4) ... infinity) \ = -ln (1-(1 / 7))-7((1 / 7^2 .2)+(1 / 7^3 .3)+(1 / 7^4 .4)+ ... . infinity) \ [ as ln (1+ x)= x-frac x^22+frac x^33-frac x^44 ... . infinity] \ [ as ln (1- x)=-( x+frac x^22+frac x^33+frac x^44 ... . infinity)] \ = -ln (6 / 7)-7(-ln (1-(1 / 7))-(1 / 7)) \ = 6 ln (6 / 7)+1 \

The integral int_0^1 \frac{1}{7^{\left[\frac{1}{x}\rig... | null (Mathematics)

\begin{aligned} & \int_0^1 \frac{1}{7^{\left[\frac{1}{x}\right]}} \mathrm{dx}=-\int_1^0 \frac{1}{7^{\left[\frac{1}{x}\right]}} \mathrm{dx} \\ = & (-1)\left[\int_1^{1 / 2} \frac{1}{7} \mathrm{dx}+\int_{1 / 2}^{1 / 3} \frac{1}{7^2} \mathrm{dx}+\int_{1 / 3}^{1 / 4} \frac{1}{7^3} \mathrm{dx}+\ldots \ldots \infty\right] \\ = & \left(\frac{1}{7}+\frac{1}{2.7^2}+\frac{1}{3.7^3}+\ldots \infty\right)-\left(\frac{1}{7.2}+\frac{1}{7^2 .3}+\frac{1}{7^2 .4} \ldots \infty\right) \\ = & -\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^2 .2}+\frac{1}{7^3 .3}+\frac{1}{7^4 .4}+\ldots . \infty\right) \\ & {\left[\text { as } \ln (1+\mathrm{x})=\mathrm{x}-\frac{\mathrm{x}^2}{2}+\frac{\mathrm{x}^3}{3}-\frac{\mathrm{x}^4}{4} \ldots . \infty\right] } \\ & {\left[\text { as } \ln (1-\mathrm{x})=-\left(\mathrm{x}+\frac{\mathrm{x}^2}{2}+\frac{\mathrm{x}^3}{3}+\frac{\mathrm{x}^4}{4} \ldots . \infty\right)\right] } \\ = & -\ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right) \\ = & 6 \ln \frac{6}{7}+1 \end{aligned}

Explanation

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