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I want to show the following (I am only really interested in real variables/parameters, so $x,b,c\in\mathbb{R}$):

$$\lim_{x\rightarrow\infty}E_1\left[b\left(x+c\right)\right]=0,\quad\text{for }b>0$$

Where I want to use the definition of the exponential integral:

$$E_1\left(x\right)=\int^{\infty}_{x}\frac{e^{-t}}{t}dt,\quad(x>0)$$

I am not sure how to show this properly. If I try to just plug in the definition, I believe I can see the expected result if I exchange the limit and the integration - but being a physicist I don't really know why I would be allowed to. I read about the dominated convergence theorem, but don't understand if/how this can be applied. Can someone give me a mathematically correct (and in the best case easy to understand) derivation?

Thanks for any hints in advance.

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Think of $E_1(n)$ for $n\ge 1$ as $$\int_1^\infty 1_{[n,\infty]}(t){e^{-t}\over t}\,dt$$ where $1_{[n,\infty]}$ denotes a function that is $1$ on $[n,\infty]$ and $0$ otherwise. The integrand for all $n\geq1$ is dominated by $e^{-t}/t$ (which is positive and has a finite integral). The integrands converge pointwise to $0$ as $n\to \infty$. The integrals thus converge to zero, and so $\lim_{n\to\infty}E_1(n)=0$. Since $E_1(x)$ is decreasing as a function of $x$ on account of the integrand being positive and since $b>0$, it follows that $\lim_{x\to\infty}E_1(b(x+c))=0$.

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  • $\begingroup$ Many thanks for your reply. > The integrands converge pointwise to 0 as $n\rightarrow\infty$. The > integrals thus converge to zero In this argumentation, you are exchanging integral and limit in $\lim_{n\rightarrow\infty}\int^{\infty}_1f_{n}\left(t\right)dt$, right? Why is this allowed? Because of the pointwise convergence? $\endgroup$ – jitter Jan 9 '16 at 13:08
  • $\begingroup$ ($f_n\left(t\right)$ being your integrand) $\endgroup$ – jitter Jan 9 '16 at 13:14
  • $\begingroup$ I made use of the dominated convergence theorem as I understood it, namely the Lebesgue theorem given, e.g., in Baby Rudin 11.32. The theorem says that if $f_n\to f$ pointwise and $|f_n|\le g$ for all $n$, with the integral of $g$ finite, then the integrals $\int f_n$ converge to $\int f$. $\endgroup$ – ForgotALot Jan 9 '16 at 17:06

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