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

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

## marked as duplicate by Guy Fsone, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 20 '17 at 0:54

• I didn't understand the 2nd step...how did you compute $I'(a)$? – SchrodingersCat Oct 30 '15 at 18:47
• Actually above we have differentiate w r. to $a$ . So above $a$ is variable and $x$ is Constant. – juantheron Oct 30 '15 at 18:51
• Sub $x=e^{-u}$ and get a Frullani integral. See math.stackexchange.com/questions/701557/… – Ron Gordon Oct 30 '15 at 18:53
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.