Let α ∈ (0,1) and β = log_e(1-α ). Let P_n(x)=x+ x^2 2+ x^3 3+… .+ x^n n,x∈ (0,1). Then the integral ∫ _0^α t^50 1-tdt is equal to

Let

$$I={\int }_{0}^{\alpha }\frac{{t}^{50}}{1-t}dt$$I=∫0α​1−tt50​dt

$$\Rightarrow I=-{\int }_{0}^{\alpha }(\frac{1-{t}^{50}-1}{1-t})dt$$⇒I=−∫0α​(1−t1−t50−1​)dt

$$\Rightarrow I=-{\int }_{0}^{\alpha }(\frac{1-{t}^{50}}{1-t})dt+{\int }_{0}^{\alpha }(\frac{1}{1-t})dt$$⇒I=−∫0α​(1−t1−t50​)dt+∫0α​(1−t1​)dt

$$\Rightarrow I=-{\int }_{0}^{\alpha }(1+t+{t}^{2}+...+{t}^{49})dt-{[\mathrm{ln}(1-t)]}_{0}^{\alpha }$$⇒I=−∫0α​(1+t+t2+...+t49)dt−[ln(1−t)]0α​

$$\Rightarrow I=-{[t+\frac{{t}^{2}}{2}+\frac{{t}^{3}}{3}+...+\frac{{t}^{50}}{50}]}_{0}^{\alpha }-\mathrm{ln}(1-\alpha )$$⇒I=−[t+2t2​+3t3​+...+50t50​]0α​−ln(1−α)

$$\Rightarrow I=-[\alpha +\frac{{\alpha }^{2}}{2}+\frac{{\alpha }^{3}}{3}+...+\frac{{\alpha }^{50}}{50}]-\mathrm{ln}(1-\alpha )$$⇒I=−[α+2α2​+3α3​+...+50α50​]−ln(1−α)

$$\Rightarrow I=-(\beta +{P}_{50}(\alpha ))$$⇒I=−(β+P50​(α))