Find $\lim_{n\rightarrow\infty}\int_{2n\pi}^{2(n+1)\pi}x\ln x\cos x\,dx$ 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?
 A: 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}
A: 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: $|(\cos x)/x |\le 1/x.$
