definite Integral with limit approches $\infty$

Question

Evaluation of $$\lim_{n\rightarrow \infty}\int^{\infty}_{0}\bigg(1+\frac{t}{n}\bigg)^{-n}\cdot \cos\bigg(\frac{t}{n}\bigg)dt$$

What i Try: put $\displaystyle \frac{t}{n}=u.$ Then $dt=ndu$

$$I_{n}=\lim_{n\rightarrow \infty}\int^{\infty}_{0}n(1+u)^{-n}\cos(u)du$$

By using by parts

$$I_{n}=\lim_{n\rightarrow \infty}n\bigg([-n(1+u)^n\cos(u)\bigg|^{\infty}_{0}+\int^{\infty}_{0}n(1+u)^{n-1}\cos(u)du\bigg]$$

How do i solve it Help me please. Thanks

I think your substitution is only going backwards in terms of problem solving progress. Keep it as is and just apply dominated convergence. — Ninad Munshi, Jul 02 '20 at 14:57
Thanks Ninad Munshi. I did not understand How can i apply DCT here. — jacky, Jul 02 '20 at 14:58
When i take limit . Then it is in $\frac{\infty}{\infty}$ form.using D.L Hopital rule. It beco.e $0$ — jacky, Jul 02 '20 at 15:04
I'm not sure what to tell you, I don't get an indeterminate form when I do it. $\cos(0) = 1$, for one thing. — Ninad Munshi, Jul 02 '20 at 15:09

score 1 · Accepted Answer · answered Jul 02 '20 at 15:23

Taking the limit inside, we get:

\begin{equation} I=\int\limits_{0}^{+\infty}\lim_{n\rightarrow \infty} \left(1+\frac{t}{n}\right)^{-n}\cos\left(\frac{t}{n}\right)\,dt \end{equation}

Let $L$ be the limit, then using the product rule, we can split the limit as follows:

\begin{equation} L=\underbrace{\lim_{n\rightarrow \infty}\left(1+\frac{t}{n}\right)^{-n}}_{L_{1}}\times\underbrace{\lim_{n\rightarrow \infty} \cos\left(\frac{t}{n}\right)\,dt}_{L_{2}} \end{equation}

The second limit is equal to $1$ and the first limit is the definition of the exponential function $e^{-t}$. Then, we conclude that $L=e^{-t}$. Plugging this into I:

\begin{equation} I=\int\limits_{0}^{+\infty}e^{-t}\,dt=\Gamma(1)=0!=1 \end{equation}

definite Integral with limit approches $\infty$

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