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.
Best Answer
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.