Does $$I=\int_{-\infty}^{\ln2}\dfrac{e^{-t}}{t^2+1} \, dt$$ converge?

First as $f(t)=\dfrac{e^{-t}}{t^2+1}$ is continuous over $\mathbb{R}$ then $f$ is Riemann integrable over $[a,b]$ where $a<b$.

$\displaystyle \int_{-\infty}^{\ln2}\dfrac{e^{-t}}{t^2+1} \, dt =\int_{-\ln2}^{+\infty}\dfrac{e^{t}}{t^2+1} \, dt $

Let $g(t):=\dfrac{e^{t}}{t^2+1}$

As $\dfrac{t}{g(t)}\underset{t\to+\infty}{\longrightarrow}0,\quad \exists a>0,\;\forall t>a\quad \dfrac{t} {g(t)}<1\iff t=\mathcal{O}\big(g(t)\big)$

We can split : $$\int_{-\ln2}^{+\infty}\dfrac{e^{t}}{t^2+1} \, dt= \int_{-\ln2}^{a}\dfrac{e^{t}}{t^2+1} \, dt+\int_{a}^{+\infty}\dfrac{e^{t}}{t^2+1} \, dt$$

So the first one converges because the integral of a Riemann-integrable function over a closed interval is finite.

But :

As $t=\mathcal{O}\big(g(t)\big)$ and $\displaystyle \int_a^{+\infty}t\;dt$ diverges $\displaystyle \implies \int_{a}^{+\infty}\dfrac{e^{t}}{t^2+1} \, dt$ diverges as well,it follows $I$ diverges. The proof is complete. $\blacksquare$

I think it is correct. I've tried to be rigorous. Is there anything better, more concise?


Why so complicated?



First of all, note that your integrand is locally bounded. Therefore all that matters is the behavior at $- \infty$. There your integrand is essentially $e^x / x^2$, $x := - t$. Removing the $1$ from the denominator doesn't hurt (you can restrict yourself e.g. to $t < - 999$ and have $\left( \frac{999}{1000} \right)^2 \frac{1}{t^2} < \frac{1}{1 + t^2} < \frac{1}{t^2}$, so whether you get divergence or convergence, the same will hold for the integral with the 1 removed).

But $e^{-t} > t^2$ for $t$ big enough. (This is probably the case for $t = -999$ and $\frac{\mathrm d}{\mathrm d (-t)} (e^{-t} - (-t)^2) = e^{-t} - 2(-t) > 0$, provided you can prove the latter. And the latter can be proved by showing $\frac{\mathrm d }{\mathrm d(-t)} (e^{-t} - 2(-t)) = e^{-t} - 2 > 0$ for $t < - 999$)

Therefore you can estimate $$\int_{-\infty}^{-999} \frac{e^{-t}}{t^2}\,\mathrm d t > \int_{-\infty}^{-999} 1\,\mathrm{d}t = \infty$$.

(You said you tried to be rigorous. My point here was to give some simple arguments to convey some feeling for what properties of the integrand matter).

[Edit: Deleted wrong remark]

  • $\begingroup$ +1 for being quite complete, and also for the last line. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 0:55
  • $\begingroup$ Instead of using $t=-999$, why not be more rigorous and use $f(x)=e^{-x}/x^2$ and find when $f'(x)=0$? $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 1:00
  • $\begingroup$ If you want to openly say a particular remark was wrong, you can use <s> ... </s>. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 1:04
  • $\begingroup$ Oh, absolutely. For a really elementary, really complete proof that should be the way to go. $\endgroup$ – jacques Aug 26 '17 at 1:12
  • $\begingroup$ Thanks for pointing out <s>! $\endgroup$ – jacques Aug 26 '17 at 1:13

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