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Some weirder cases of improper integrals and probably the worse way to evaluate them.

( sorry im posting 3 even though they are pretty tough but there all very related and half my problem is how to layout my solution)

1) $\int^{\infty}_{0} \frac {x^{(1/2)}}{e^{x}-1}$

(i) this one i had to go all cheese no skill first clearly at infinity this function is dominated by ${e^{-x}}$ as $x \to \infty$ F(x)=0 by the comparison test F(x) converges at $\infty$ the real problem happens at 0 as we have $\frac{0}{0}$ i used a taylor expansion here to the first variable thus $e^{x} \approx 1+x$ which leads us to $\frac {x^{(1/2)}}{1-x-1}$=$\frac {1}{x^{1/2}}$ as $x \to 0$ limit converges thus $\int^{\infty}_{0} \frac {x^{(1/2)}}{e^{x}-1}$ converges.

2)$\int^{\infty}_{0} \frac{\sin(\frac{1}{x})}{x^{1/5}}$ <-- thats (1/x) on top and $(x^{1/5})$ on the bottom in case you left your magnifine glass back in the 90's

(ii) firstly as $x \to 0$ we have $\sin (\frac {1}{x}) \to [-1,1]$ and everything in between. thus the funtion is clearly dominated by $\frac {1}{x^{1/5}}$ and this converges as x goes to 0. ( this part was super sketchy to me because the function gets there in a funny little hop kinda way)

more interesting is how to tackle the limit as $x \to \infty$ here i did a change of variables let u=$\frac {1}{x}$ thus $du=-(\frac {1}{x^{2}})$ we have -$(\frac {1}{x})^{2}$ and $(\frac {1}{x})^{1/5}$ which leads to $u^{11/5} \sin(u)$ since as x goes to infinity $\frac{1}{x} \to 0$

we have $\lim_{u \to 0} u^{11/5} \sin(u)$ clearly $\sin(u)$ is dominated by $u^{11/5}$

thus we have $\lim_{u \to 0} u^{11/5}=0$ thus the limit converges on both ends.

Biggest problem with above is that it seems like it shouldn't work; but it also has the problem of i have no idea what rules i used to show this ( assuming that this is even a plausible answer)

3) $\int^{\infty}_{-\infty} \frac {e^{x}}{e^{x}+x^{2}}$

(iii) this one has given me alot of trouble a taylor series expansion can't be used for obvious reason's at $\infty$ i tried a funny method of factoring but that didnt yield much that made sense lastly i tried divideing by $e^{x}$. this yields $\int^{\infty}_{-\infty} \frac {1}{e^{x}+x^{2}}$ which clearly the limit as $x \to \infty$=0 and the limit $x \to -\infty$=0 but $e^{x}$ as $x \to \infty$ clearly diverges.

i have tried computing directly $\lim_{x \to \infty} \frac {e^{x}}{e^{x}+x^{2}} =1$ but the other side doesnt lend itself well to this aproach.

Basically what i want is criticism on the three proofs on how to make them acceptable as a correct answer and also how to prove that #3 diverges ( as i am fairly certain it does i just can't pin down why) thanks for the help.

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I am not the downvoter, but I would like to mention that while I appreciate you showing your work, the question feels very "cluttered," and the fact that you are asking $3$ questions instead of just one exacerbates this problem. –  Andrew Salmon Apr 2 '13 at 5:54
    
Im not going to try and edit it again my browser is half crashing trying to load all the latex as i edit it if people feel its not helpful i will happily delete the post but not until i understand part 3 –  Faust7 Apr 2 '13 at 6:01

1 Answer 1

up vote 1 down vote accepted

It is usually good to break up an integral into parts, so that each part has at most one "bad" feature.

For your third integral, break up into the integrals from $-\infty$ to $0$, and from $0$ to $\infty$.

Let's look at $\displaystyle \int_0^\infty \frac{e^x}{e^x+x^2}\,dx$. Informally, for large $x$, the $x^2$ term is utterly negligible in comparison with $x^2$. So for big $x$ our function is nearly $1$, and of course the integral will blow up.

To be more formal, divide top and bottom of our integrand by $e^x$. We get $$\dfrac{1}{1+\frac{x^2}{e^x}}.$$ Since $\lim_{x\to\infty} \frac{x^2}{e^x}=0$, for large enough $x$ we have $\frac{x^2}{e^x}\lt \frac{1}{2}$. So for large enough $x$, the integrand is $\gt \frac{2}{3}$.

The integral $\int_0^\infty \frac{2}{3}\,dx$ clearly diverges, so by Comparison so does $\displaystyle\int_0^\infty \frac{e^x}{e^x+x^2}\,dx$.

For your second integral, break up as you did. The ideas ou used are fine. We will prove that both integrals converge absolutely, meaning that if we take absolute values we still have convergence. By a standard theorem, absolute convergence implies convergence.

For the integral from $0$ to $1$, note that $|\sin(1/x)|\le 1$. Thus $$\left|\frac{\sin(1/x)}{x^{1/5}}\right|\le \frac{1}{x^{1/5}}.$$ By a standard theorem, $\int_0^1\frac{dx}{x^{1/5}}$ converges, so by Comparison our integral from $0$ to $1$ convrges absolutely, and hence converges.

For the integral from $1$ to $\infty$, I would suggest using the inequality $|\sin t|\le |t|$. For $x\ge 1$, this implies that $0\lt \sin(\pi/x)\lt \frac{\pi}{x}$.

Thus in our interval, our function is between $0$ and $\frac{\pi}{x^{6/5}}$. The rest is done by Comparison.

For the first integral, again the intuition is good. There could be improvement in the detail.

Break up as the integral from $0$ to $1$, plus the integral from $1$ to\infty$.

For the integral from $1$ to $\infty$, true, the $e^x$ term at the bottom will cush the top. Maybe you could use the Limit Comparison Test, comparing with $\frac{1}{e^{x/2}}$.

Or else observe that for large enough $x$ (and it happens pretty early) the bottom is bigger than $(1/2)e^{x}$. for large $x$, the top is $\lt e^{x/2}$. So the ratio, for large $x$, is $\lt \frac{2}{e^{x/2}}$.

For the part from $0$ to $1$, use the fact that $\lim_{x\to 0}\frac{e^x-1}{x}=1$, or more or less equivalently use the MacLaurin series for $e^x$ to get an idea of the size of $e^x-1$.

For $x$ close enough to $0$, $x/2\le e^x-1$. so our integrand is $\le \frac{2}{x^{1/2}}$, and by Comparison with a standard integral we have convergence.

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i don't understand the line so for large enough x, the integrand >2/3 i agree that there are value for x where the integrand is greater then $2/3$ but why r u choosing this? why can you choose this? –  Faust7 Apr 2 '13 at 5:58
    
Because $x^2/e^x\lt 1/2$, the bottom is $\lt 3/2$, so the whole thing is $\gt 2/3$. –  André Nicolas Apr 2 '13 at 6:05
    
that really weird but makes some sense ( the origional setup is exactly what i tried first i just couldnt wrap my head around a bound) as for the second integral that makes alot of sense! out curiosity does what i did make any sense? –  Faust7 Apr 2 '13 at 6:08
    
First the bad news. On the third problem, as described, it did not make sense. The stuff you did on the first two problems made complete sense, and was fundamentally right. Your intuition there was reliable, and well described. The formal arguments were not tight enough. –  André Nicolas Apr 2 '13 at 6:26
    
Awesome! ( i knew #3 wasn't right the way i described it) the formula part always comes last for me; being a physics not a math student but im sure i can hammer that out with practice. Getting things out of my head and having someone put everything in the right place really solidify's my intuition. Thank you very much for all your help i am sure i won't have a problem killing anyone of these on my final. –  Faust7 Apr 2 '13 at 6:42

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