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

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?

$$\lim_{t\to-\infty}\frac{e^{-t}}{t^2+1}=+\infty$$

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]

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