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:

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

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


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

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

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.

  • $\begingroup$ Thank you! I did a noob thing.. $\endgroup$ – KYHSGeekCode Sep 23 '18 at 16:26
  • $\begingroup$ You are welcome. And don't worry, everybody does noob things from time to time... $\endgroup$ – Ernie060 Sep 23 '18 at 16:28
  • $\begingroup$ So is answer $\ln 2$? $\endgroup$ – KYHSGeekCode Sep 24 '18 at 2:41
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Ernie060 Sep 24 '18 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.