Determine the value of a limit involving an integral

Question

For any natural number $ a $, determine the value of $$ \lim_{n \to \infty} n \int_1^e x^a (\log (x))^n \,dx.$$

How we to find the value of the integral value? Please help me. Thank you in advance.

Answer 1

Consider a slightly different integral, $$I(a) = \int_{0}^ex^adx$$ Using Feynman's trick for differentiation under the integral sign, $$I^{(1)}(a) = \int_0^ex^a\log(x)dx$$ If we continue, you will see that $$I^{(n)}(a)=\int_0^ex^a(log(x))^ndx$$ We can easily compute $I(a)$ $$I(a)=\frac{e^{a+1}}{a+1}$$ Now if we write the power series expansion for $e^{a+1}$ $$I(a) = \frac{1}{a+1}+1+\frac{(1+a)}{2!}+\frac{(1+a)^2}{3!}+...$$ If we now compute the $n^{th}$ derivative, all the terms upto the power of n will vanish $$I^{(n)}(a) = \frac{n!}{(n+1)!}+\frac{(n+1)(n)(n-1)...2(1+a)}{(n+2)!}+...$$ Multiply the above expression by n and finally take the limit. Can you take it from here?

Hope this helps

Answer 2

The integral can be expressed as $$I_n{=\int_{0}^1 u^ne^{au}du }$$By the technique of integration by parts we obtain $$I_n{=\int_{0}^1 u^ne^{au}du=u^n{e^{au}\over a}\Bigg|_{0}^1-{n\over a}\int_{-\infty}^1 u^{n-1}e^{au}du\\={e^a\over a}-{n\over a}I_{n-1}}$$Note that $\lim_{n\to \infty} I_n=0$ because $u^n\to 0$ for $u\in[0,1]$, therefore we can say $$\lim_{n\to \infty}{e^a\over a}-{n\over a}I_{n-1}=0$$from which we easily conclude$$\lim_{n \to \infty} n \int_1^e x^a (\log (x))^n \,dx=e^a$$