Find the Exact value of $\lim_{n\to\infty} \sum_{k=1}^n \int_0^{\pi} e^{-nx}\sin (kx) dx$

Question

Find the Exact value of : $$\lim_{n\to\infty} \sum_{k=1}^n \int_0^{\pi} e^{-nx}\sin (kx) dx$$

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What I did :

Take $I=\int_0^{\pi}e^{-nx}\sin(kx)dx$

After Integration, I get $$I=\frac{k(1-e^{-n\pi}\cos(k\pi))}{n^2+k^2}$$

But that's the end.

I don't have any ideas to sqeeze this, And I can't change this to integrals because of $n$ which stand on exponential.

Did I miss something during calculating $I$?

Answer 1

Hint.

So you have found $$I_k=\frac{k}{k^2+n^2}-e^{-n\pi}\cdot \frac{\cos(k\pi)}{k^2+n^2}$$ The first one can be written as $$\frac{k}{k^2+n^2}=\frac{1}{n}\cdot\frac{k/n}{1+(k/n)^2}$$ which gives you a Riemann sum after taking the sum.

Bound the second term as follows $$\left|e^{-n\pi}\frac{\cos(k\pi)}{k^2+n^2}\right|\leq e^{-n\pi}$$ which gives you what after taking the sum...? Can you conclude?