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Given that α ≥ 0 and value of integral at α = 0 is 0. The value of

$\int_{0}^{1}\frac{x^{\alpha }-1}{log x}dx$

2 Answers

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Using Leibnitz Theorem,  

$\frac{\mathrm{d}I }{\mathrm{d} a}$= $\frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1} \frac{x^{\alpha }-1}{log x}dx$
= $\int_{0}^{1}\frac{\partial }{\partial a} \frac{x^{\alpha }-1}{log x}dx$ (differentiation of Integral function is equal partial differentiation of that function)
= $\int_{0}^{1} \frac{x^{a} log x}{log x}$
= $\int_{0}^{1} x^{a} dx$
= $\frac{x^{a+1}}{a+1}$ 
= $\frac{{1}}{a+1}$ 
 $\int dI$ =$\int \frac{1}{a+1} da$

= ln(a+1) + c

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I = ∫( X^a -1)/log(x) dx

Use Leibnitz theorem

dI/da =∫ (X^a) log(x)/log(x)=∫ (X^a)=(X^a+1)/a+1

put the limit and get the ans as 1/a+1  

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