0

Suppose $0<b<1$ and we are asked to evaluate $$I=\int^\infty_{0}\dfrac{x^b\ln x}{1+x}\,dx$$

Initially, I thought of using the contour in this link. But then the problem comes when I am trying to consider $C_R$. That is, $$|\int_{C_R}f(z)\,dz|\leq\pi R\dfrac{R^b\ln R}{R-1}$$ but this integral does not converge to $0$ as $R\rightarrow\infty$. How can I solve this question?

Edited: I also tried a keyhole contour as well but the problem remains the same. That is, $$|\int_{C_R}f(z)\,dz|\leq2\pi R\dfrac{R^b\ln R}{R-1}$$

Alvin
  • 7

1 Answers1

3

In this AOPS thread it's explained that for $\int_0^{\infty} \frac{x^{\alpha}}{1+x^{\beta}}\mathrm{d}x$ to converge we require $\beta > \alpha +1 >0$. Moreover, in the answers to this question it's shown that $$ \int_0^{\infty} \frac{x^{\alpha}}{1+x^{\beta}}\mathrm{d}x = \frac{\pi}{\beta}\csc\left(\frac{\pi(\alpha+1)}{\beta} \right) $$ whenever the parameters $\alpha, \beta$ are in the previously stated convergence range. Now, differentiating the previous equation with respect to $\alpha$ gives $$ \boxed{\int_0^{\infty} \frac{x^{\alpha}\ln(x)}{1+x^{\beta}}\mathrm{d}x = -\frac{\pi^2}{\beta^2}\csc\left(\frac{\pi(\alpha+1)}{\beta} \right)\cot\left(\frac{\pi(\alpha+1)}{\beta} \right), \qquad \beta > \alpha +1 >0} $$

If you're certain that the condition $0<b<1$ is correct then there was probably a mistake in the denominator (maybe it was meant to be $1+x^{\color{red}{2}}$ instead of $1+x$). Otherwise, as repeatedly pointed out in the comments, $\int_0^{\infty} \frac{x^{b}\ln(x)}{1+x}\mathrm{d}x $ only converges for $-1<b<0$, and as such your integral (as currently stated) has no evaluation other than "Divergent".

Robert Lee
  • 7,233