# Testing whether the following integral converges

Determine whether the following integral converges or diverges: $$I=\int_0^\infty\frac{x^{80}+\sin(x)}{\exp(x)}\,dx.$$

Since $$0\overset{?}{<}\frac{x^{80}+\sin(x)}{e^x}\leq\frac{x^{80}+1}{e^x}\sim\frac{x^{80}}{e^x}$$

and if we let $$f(x)=x^{80}/e^x$$ and $$g(x)= \sqrt{x}\,e^{-\sqrt{x}}\,$$ we have that $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$ therefore if $$\int_0^\infty g(x)\,dx$$ converges so does $$I$$. But $$\int_0^\infty\sqrt{x}\,e^{-\sqrt{x}}\,dx\to {\small{\begin{bmatrix}&u=\sqrt{x}&\\&du=\frac{1}{2\sqrt{x}}dx&\end{bmatrix}}} \to\int_0^\infty e^{-u}\,du=1.$$ Thus $$I$$ does converge.

Now, the limit used above is not too trivial to show that it is $$0$$. Intuitively, it is easy to see why, but on a more rigorous aspect, L'Hôpital's Rule here would get messy. Thus, my question, is there a more straightforward way to show that the above integral converges?

• Better if you compare with $g(x)=e^{-x/2}$. Commented Nov 12, 2018 at 22:26
• Yes you are right. No idea how I missed that. Thank you! Commented Nov 12, 2018 at 22:30
• Also, the substitution you made wasn't right. Actually, you should have got $$\int_0^\infty 2u^2e^{-u}du,$$ (which is convergent indeed, since it's equal to $2\Gamma(3)=2\cdot 2!=4$.) Commented Nov 12, 2018 at 22:35
• $$\int_0^\infty (x^{80}e^{-x}+\sin(x)e^{-x})\,dx=80! +0.5$$ Commented Nov 12, 2018 at 22:38
• L'Hopital's rule is simply enough with $f(x)=x^{80}$ and $g(x)=e^x$. Keep taking derivatives and the numerator ends up as a constant $80!$, while the denominator remains $e^x$. The integral is obviously convergent. Commented Nov 12, 2018 at 22:44

The term $$\sin x\,e^{-x}$$ obviously converges.

Now by the Taylor development,

$$e^x>\frac{x^{82}}{82!}$$ and

$$x^{80}e^{-x}<82!\,x^{-2},$$ which has a convergent integral.

Is there a more straightforward way to show that the above integral converges?

We can actually evaluate the integral of interest. Note that we have

\begin{align} \int_0^\infty (x^{80}+\sin(x))e^{-x}\,dx&=\int_0^\infty x^{80}e^{-x}\,dx+ \int_0^\infty \sin(x)e^{-x}\,dx\\\\ &\Gamma(81)+\frac12\\\\ &=80!+0.5 \end{align}

There is no need for L'Hopital's Rule here. To show that $$\int_0^{\infty} \frac {x^{80}} {e^{x}}\, dx <\infty$$ note that $$\frac {x^{80}} {e^{x}} or $$x^{80} for $$x$$ sufficiently large: $$e^{x/2} > \frac {(x/2)^{100}} {100!} >x^{80}$$ if $$x^{20}>\frac {2^{100}} {100!}$$. Since $$e^{-x/2}$$ is integrable we are done.

As was pointed out in the comments, you could have simply use $$e^{x/2}$$. However even the function you have used can be proved in the limit. Start with dividing $$f(x)$$ by $$g(x)$$

$$\frac{f(x)}{g(x)} = \frac{x^{79.5}}{e^{\sqrt{x}}}$$

Now, generally we know that for any $$a$$, the limit $$x^a/e^x$$ as x tends to infinty is $$0$$. In this case, however, we have $$e^{\sqrt{x}}$$. Now if you start applying the $$L'Hôpital's Rule$$, you will see that this limit also degrades albeit at only half the exponential rate i.e. the numerator decreases by $$\sqrt{x}$$ every time.

For example, applying the rule for the first time will give you

$$2(79.5)\frac{x^{79.5 - 1}\sqrt{x}}{e^{\sqrt{x}}} = 2(79.5)\frac{x^{79}}{e^{\sqrt{x}}}$$

Here the numerator decreased by $$0.5$$ power of x whereas the denominator remained the same. So if you apply the rule sufficiently enough number of times, we'll get the limit as $$0$$.