Find $\lim_{n\rightarrow\infty}\int_{2n\pi}^{2(n+1)\pi}x\ln x\cos x\,dx$

Question

Find $$\lim_{n\rightarrow\infty}\int_{2n\pi}^{2(n+1)\pi}x\ln x\cos x\,dx$$. Integrating by parts I obtained that the integral is equal to $$-\int_{2n\pi}^{2(n+1)\pi}\ln x \sin x\, dx$$. Integrating again by parts, I get $$-\int_{2n\pi}^{2(n+1)\pi}\ln x\sin x\,dx=\ln\frac{2(n+1)\pi}{2n\pi}-\int_{2n\pi}^{2(n+1)\pi}\frac{\cos x}{x}dx$$. I know that $$|\frac{\cos x}{x}|\leq1$$ and by integrating it I get that $$\int_{2n\pi}^{2(n+1)\pi}\frac{\cos x}{x}\in(-2\pi,2\pi)$$. But this does not provide me with a result. Any help?

Answer 1

NOT A SOLUTION: This may or may not be of help. I hope it's of use.

Here we will address your limit by fist addressing the integral: \begin{equation} I = \int_{2n\pi}^{2(n + 1)\pi}x \ln(x) \cos(x)\:dx\nonumber \end{equation} We first observe that: \begin{equation} \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a} x^a =\ln(x)\nonumber \end{equation} Thus (and by the Dominated Convergence Theorem and Leibniz's Integral Rule $I$ becomes: \begin{equation} I = \int_{2n\pi}^{2(n + 1)\pi}x \left( \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a} x^a \right) \cos(x)\:dx = \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a} \int_{2n\pi}^{2(n + 1)\pi}x^{a + 1}\cos(x)\:dx\nonumber \end{equation} Here we make the substitution $x = t + 2n\pi$: \begin{equation} I = \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a} \int_0^{2\pi} \left(t + 2n\pi\right)^{a + 1} \cos\left(t + 2n\pi \right)\:dt \end{equation} Now $\cos\left(t + 2n\pi \right) = \cos\left(t\right)$. Thus: \begin{equation} I = \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a} \int_0^{2\pi} \left(t + 2n\pi\right)^{a + 1} \cos\left(t\right)\:dt\nonumber \end{equation}

In terms of a closed form, I am unsure how to continue (if the upper limit of the integral was infinity it would be easy!) Of we now incorporate the limit (I will call the result $J$): \begin{equation} J = \lim_{n \rightarrow \infty} \lim_{a \rightarrow 0^+}\frac{\partial}{\partial a}\int_0^{2\pi} \left(t + 2n\pi\right)^{a + 1} \cos\left(t\right)\:dt\nonumber \end{equation}

Answer 2

You obtained $$\ln\frac{2(n+1)\pi}{2n\pi}-\int_{2n\pi}^{2(n+1)\pi}\frac{\cos x}{x}dx.$$

You are further along than you think. In the first expression, you are applying $\ln$ to an expression that has limit $1.$ In the integral, you gave away the ranch. You wrote $|(\cos x)/x |\le 1$ but much more is true: $|(\cos x)/x |\le 1/x.$