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This question already has an answer here:

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

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marked as duplicate by Guy Fsone, Namaste calculus Dec 20 '17 at 0:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I didn't understand the 2nd step...how did you compute $I'(a)$? $\endgroup$ – SchrodingersCat Oct 30 '15 at 18:47
  • $\begingroup$ Actually above we have differentiate w r. to $a$ . So above $a$ is variable and $x$ is Constant. $\endgroup$ – juantheron Oct 30 '15 at 18:51
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    $\begingroup$ Sub $x=e^{-u}$ and get a Frullani integral. See math.stackexchange.com/questions/701557/… $\endgroup$ – Ron Gordon Oct 30 '15 at 18:53
  • $\begingroup$ Looks OK to me; also, it looks quite natural to solve it this way. As to any other method... well, surely some must be possible, but why bother. Any other answer would be either wrong or the same as this one. $\endgroup$ – Ivan Neretin Oct 30 '15 at 18:53
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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.

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