Evaluation of $\int_{0}^{1}\frac{x^{2015}-1}{\ln x}dx$

Question

Evaluation of $\displaystyle \int_{0}^{1}\frac{x^{2015}-1}{\ln x}dx\;\;$

$\bf{My\; Try::}$ Let $$I(a) = \int_{0}^{1}\frac{x^{a}-1}{\ln x}dx\;,$$ Then $$I'(a) = \int_{0}^{1}\frac{x^a\cdot \ln(x)}{\ln(x)}dx = \int_{0}^{1}x^{a}dx = \left[\frac{x^{a+1}}{a+1}\right]_{0}^{1}=\frac{1}{a+1}$$

So we get $$I(a) = \ln|a+1|+\mathcal{C}.$$

Now When $a=0\;,$ We get $I(0) =0$

So we get $I(0)=\ln(1)+\mathcal{C}\Rightarrow 0 = 0+\mathcal{C}\Rightarrow \mathcal{C}=0$

So we get $$I(a) = \int_{0}^{1}\frac{x^{a}-1}{\ln(x)}dx = \ln|a+1|$$

So $$I(2015) = \int_{0}^{1}\frac{x^{2015}-1}{\ln(x)}dx = \ln|2016|$$

can we solve it by using any other Method Like Using Double Integration.

If yes Then plz explain here, Thanks

Answer 1

One may consider the double integral $$ I:=\int_0^1\!\!\int_0^1a\:x^{ay} dx\:dy,\qquad a\geq0. $$ Applying Fubini's theorem, on the one hand, we have $$ I=\int_0^1\!\left(\int_0^1a\:x^{ay} dy\right)dx=\int_0^1\frac{x^a-1}{\ln x}\:dx. $$ On the other hand,$$ I=\int_0^1\!\left(\int_0^1a\:x^{ay} dx\right)dy= \int_0^1\!\!\frac{a}{ay+1}dy=\int_0^1\frac{(ay+1)'}{ay+1}dy=\ln(a+1). $$ Thus

$$ \int_0^1\frac{x^a-1}{\ln x}dx=\ln(a+1),\qquad a\geq0. $$

Then we put $a:=2015$ to obtain the initial integral.