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We're told that $$f_n(\lambda)=\int_{n}^{+\infty} e^{-\lambda t}\sin{t}\ dt$$ converges uniformly to $0$ when $n\to\infty$.

Is it really true ? I can't see how to bound that with something going to $0$ when $n\to\infty$ independently of $\lambda$...

For example, we could do

$$\left| \int_{n}^{+\infty} e^{-\lambda t}\sin{t}\ dt\right|\le \int_{n}^{+\infty} e^{-\lambda t}\ dt=\frac{e^{-\lambda n}}{\lambda}\xrightarrow{n\to+\infty} \ 0,$$ but the convergence isn't uniform, is it ?

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What do you mean by uniform convergence of a sequence of numbers? – Brian M. Scott Oct 1 '11 at 9:43
Say the integral above is $f_n(\lambda)$. I mean that $f_n\to 0$ uniformly. – Klaus Oct 1 '11 at 9:47
up vote 6 down vote accepted

Writing $\sin(t)$ as the imaginary part of $\mathrm e^{it}$ and computing a primitive of the complex exponential $\mathrm e^{-(\lambda-i)t}$, one gets $$ f_n(\lambda)=\mathrm e^{-\lambda n}\,\frac{\lambda\sin(n)+\cos(n)}{\lambda^2+1}. $$ One sees that $f_n(0)=\cos(n)$. Since $\cos(n)$ does not converge to zero, $f_n$ does not converge to zero uniformly on $\lambda\geqslant 0$.

The same objection applies to the uniform convergence on $\lambda>0$, choosing $\lambda=\frac1n$.

But $|f_n(\lambda)|\leqslant\mathrm e^{-\lambda n}$ hence $f_n$ converges to zero uniformly on $\lambda\geqslant\lambda_0$ for every positive $\lambda_0$.

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First, the integral who defines $f_n$ is convergent for $\lambda>0$. The inequality $|f_n(\lambda)|\leq \frac{e^{-\lambda n}}{\lambda}$ show that we have the uniform convergence on each interval $\left[a,+\infty\right($, where $a>0$. But we can't get more. Indeed \begin{align*} f_n\left(\frac 1n\right)&=\int_n^{+\infty}e^{-\frac tn}\sin tdt\\ &=\left[e^{-\frac tn}(-\cos t)\right]_n^{+\infty}-\int_n^{+\infty}e^{-\frac tn}\left(-\frac 1n\right)(-\cos t)dt\\ &=e^{-1}\cos n-\frac 1n\int_n^{+\infty}e^{-\frac tn}\cos tdt\\ &=e^{-1}\cos n-\frac 1n\left(\left[e^{-\frac tn}\sin t\right]_n^{+\infty}-\int_n^{+\infty}e^{-\frac tn}\left(-\frac 1n\right)\sin tdt\right)\\ &=e^{-1}\cos n+\frac{e^{-1}\sin n}n-\frac 1{n^2}\int_n^{+\infty}e^{-\frac tn}\sin tdt\\ &=e^{-1}\cos n+\frac{e^{-1}\sin n}n-\frac 1{n}\int_1^{+\infty}e^{-u}\sin (nu)du, \end{align*} and $\displaystyle\lim_{n\to +\infty}\frac{e^{-1}\sin n}n-\frac 1{n}\int_1^{+\infty}e^{-u}\sin (nu)du=0$. If the convergence was uniform to $0$ on $\mathbb R^*_+$, we would have $\displaystyle\lim_{n\to +\infty}f_n\left(\frac 1n\right)=0$, but it can't be the case since the sequence $\{\cos n\}$ is not convergent.

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