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I want to find the value of $$\int_0^1 \frac{x-1}{\log x}\,dx$$ where $\log x$ stands for the natrual logarithm.

I put it in the wolfram alpha, and it saids it's $\log 2$. (Refer to : http://www.wolframalpha.com/input/?i=int_0%5E1+%28x-1%29%2F%28lnx%29)

But I can't figure out how it comes to that value. Help me to find out the procedure from that integration to get $\log 2$.

Matholic
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  • And sorry for the inconvenience of ill-written formula, could you tell me where the syntax for Mathjax is wrong? – Matholic Dec 27 '14 at 17:47
  • You need to clarify the question. It is not at all clear what the formula is meant to be... Did you mean $\int_0^1 \frac{x-1}{\ln x} dx$? To see how I wrote the latter, right click over the formula and select 'Show Math as' and select 'Tex Commands'. – copper.hat Dec 27 '14 at 17:47
  • Thanks to @copper.hat, that's what I wanted to ask for. – Matholic Dec 27 '14 at 17:50
  • But in your link the function is $x-1/\ln(x)$. You first have to learn the basics of writing a fraction of functions before asking questions. – Karl Dec 27 '14 at 17:58
  • @copper.hat I see there $\int (x-1/\ln(x))dx$ and not $\int ((x-1)/\ln(x))dx$ as asked. – Karl Dec 27 '14 at 18:01
  • @copper.hat just did it, the same. – Karl Dec 27 '14 at 18:05
  • @Karl: Dunno what to say, it looks fine to me when I click through... – copper.hat Dec 27 '14 at 18:06

1 Answers1

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Here's a method using the Feynman integration trick: Consider $$I(a)=\int\limits_0^1 \frac{x^a-1}{\log x}\,dx$$

Then we have $$I'(a)=\frac{d}{da}\int\limits_0^1 \frac{x^a-1}{\log x}\,dx=\int\limits_0^1 \frac{d}{da}\left[\frac{x^a-1}{\log x}\right]\,dx=\int\limits_0^1 x^a\,dx=\frac{1}{a+1}$$

Now integrating with respect to $a$ gives $$I(a)=\log(a+1)+C$$

Now, to find $C$, we can note that the $I(0)=0$ (because when $a=0$ the numerator turns into $1-1=0$). Hence $0=\log(0+1)+C$, so $C=0$.

Hence we have $I(a)=\log(a+1)$, and in particular we care about the $a=1$ case in which the solution is just $\log(2)$.

Peter Woolfitt
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