# Evaluating $\int^1_0 \frac{x-1}{\ln x}dx$ using Leibniz's integral formula

I read a wikipedia article about Leibniz integral rule and its applications.

It stated that $\int^1_0 \frac{x-1}{\ln x}dx$ can be evaluated with this rule, so I tried it.

I started by:

$$f(\alpha)=\int^1_0\frac{x^\alpha-1}{\ln x}dx$$

$$\int^1_0 \frac{x-1}{\ln x}dx=f(1)$$

$$f(0)=0$$

$$f'(\alpha)=\int^1_0\frac{\partial}{\partial \alpha}\frac{x^\alpha-1}{\ln x} dx$$

$$=\ln \alpha \int^1_0 \frac{x^\alpha}{\ln x}dx$$

I was stuck here, and tried:

$$f''(\alpha)=\frac 1\alpha \int^1_0\frac{x^\alpha}{\ln x}dx+ {(\ln \alpha)}^2\int^1_0\frac{x^\alpha}{\ln x}dx$$

$$=\left(\frac 1\alpha + {(\ln \alpha)}^2\right)f'(\alpha)$$

$$\therefore \frac{f''(\alpha)}{f'(\alpha)}=\left(\frac 1\alpha + {(\ln \alpha)}^2\right)$$

$$\therefore \ln |f'(\alpha)|=\ln \alpha +\alpha {(\ln\alpha)}^2-2\alpha\ln\alpha+2\alpha+C$$

$$|f'(\alpha)|=\alpha e^{\alpha {(\ln\alpha)}^2-2\alpha\ln\alpha+2\alpha+C}$$

Stuck again!

How can I evaluate $\int^1_0 \frac{x-1}{\ln x}dx$ using Leibniz's integral formula?

Thank you.

You made a mistake in the derivative under the integral. The derivative is $$\frac{\partial}{\partial \alpha}x^\alpha = x^\alpha \ln x$$, not $$x^\alpha \ln \alpha$$. The integral then becomes $$f'(\alpha) = \int_0^1 x^\alpha \, dx$$ which is much easier.
• So is answer $\ln 2$? – KYHSGeekCode Sep 24 '18 at 2:41