## 175 – Quiz 6

Here is quiz 6.

Problem 1 is Exercise 7.7.64 from the book. It asks to determine the convergence of ${\displaystyle\int_{-\infty}^\infty \frac{dx}{e^x+e^{-x}}.}$

There are several ways of approaching this problem.

Method 1: Begin by noting that

$\displaystyle \frac1{e^x+e^{-x}}=\frac{e^x}{e^{2x}+1},$

and so

$\displaystyle \int \frac{dx}{e^x+e^{-x}}=\int\frac{e^x}{e^{2x}+1}\,dx$

can be evaluated by using the substitution ${u=e^x,}$ so ${du=e^x\,dx,}$ which transforms the integral into

$\displaystyle \int\frac{du}{u^2+1},$

that can be solved with the trig. substitution ${u=\tan\theta,}$ so ${du=\sec^2\theta\,d\theta,}$ and

$\displaystyle \int\frac{du}{u^2+1}=\int d\theta=\theta+C=\tan^{-1}(e^x)+C.$

$\displaystyle \begin{array}{rl} \displaystyle \int_{-\infty}^\infty\frac{dx}{e^x+e^{-x}}&=\displaystyle \lim_{a\rightarrow-\infty}\int_a^0\frac{dx}{e^x+e^{-x}}+\lim_{b\rightarrow\infty}\int_0^b\frac{dx}{e^x+e^{-x}}\\ &=\displaystyle \lim_{a\rightarrow-\infty}\left.\tan^{-1}(e^x)\right|_a^0+\lim_{b\rightarrow\infty}\left.\tan^{-1}(e^x)\right|_0^b\\ &=\displaystyle\left(\frac{\pi}4-(-\frac{\pi}2)\right)+\left(\frac{\pi}2-\frac{\pi}4\right)\\ &=\pi. \end{array}$

We have found the value of the integral, so in particular, it converges.

Method 2: First,

$\displaystyle \int_{-\infty}^\infty\frac{dx}{e^x+e^{-x}}=\lim_{a\rightarrow-\infty}\int_a^0\frac{dx}{e^x+e^{-x}}+\lim_{b\rightarrow\infty}\int_0^b\frac{dx}{e^x+e^{-x}}.$

We use comparison for both ${x\in(-\infty,0)}$ and ${x\in(\infty,0):}$

If ${x<0,}$ ${\displaystyle\frac1{e^x+e^{-x}}<\frac1{e^{-x}}=e^x.}$ Since

$\displaystyle \lim_{a\rightarrow-\infty}\int_a^0e^x\,dx=\lim_{a\rightarrow-\infty}\left.e^x\right|_a^0=1,$

we have by comparison that ${\displaystyle\int_{-\infty}^0\frac{dx}{e^x+e^{-x}}}$ is finite.

If ${x>0,}$ ${\displaystyle\frac1{e^x+e^{-x}}<\frac1{e^{x}}=e^{-x}.}$ Since

$\displaystyle \lim_{b\rightarrow\infty}\int_0^be^{-x}\,dx=\lim_{b\rightarrow\infty}\left.-e^{-x}\right|_0^b=1,$

we have by comparison that ${\displaystyle\int_0^{\infty}\frac{dx}{e^x+e^{-x}}}$ is also finite.

Hence, ${\displaystyle\int_{-\infty}^\infty\frac{dx}{e^x+e^{-x}}}$ converges.

Method 3: As before,

$\displaystyle \int_{-\infty}^\infty\frac{dx}{e^x+e^{-x}}=\lim_{a\rightarrow-\infty}\int_a^0\frac{dx}{e^x+e^{-x}}+\lim_{b\rightarrow\infty}\int_0^b\frac{dx}{e^x+e^{-x}}.$

The change of variables ${u=-x}$ transforms ${\displaystyle\int_a^0\frac{dx}{e^x+e^{-x}}}$ into

$\displaystyle \int_{-a}^0\frac{-du}{e^{-u}+e^u}=\int_0^{-a}\frac{du}{e^{-u}+e^u},$

and

$\displaystyle \lim_{a\rightarrow-\infty}\int_0^{-a}\frac{du}{e^{-u}+e^u}=\int_0^\infty\frac{du}{e^{-u}+e^u},$

which of course is the same as ${\displaystyle\int_0^\infty\frac{dx}{e^x+e^{-x}}.}$

So all we need to do is to show that ${\displaystyle\int_0^\infty\frac{dx}{e^x+e^{-x}}}$ converges, and for this we can use comparison, because

$\displaystyle \frac1{e^x+e^{-x}}<\frac1{e^x},$

and

$\displaystyle \int_0^\infty\frac{dx}{e^x}=\lim_{b\rightarrow\infty}\int_0^be^{-x}\,dx=\lim_{b\rightarrow\infty}\left.-e^{-x}\right|_0^b=1.$

Problem 2 is an easier version of Exercise 7.7.63 from the book. It asks to determine the convergence of ${\displaystyle\int_0^\infty\frac{dx}{\sqrt{x^4+1}}.}$

The problem here is that we cannot find ${\displaystyle\int\frac{dx}{\sqrt{x^4+1}},}$ so we are forced to use comparison. A natural thing to try would be to compare ${1/\sqrt{x^4+1}}$ with ${1/\sqrt{x^4}=1/x^2.}$

Since ${x^2=\sqrt{x^4}<\sqrt{x^4+1},}$ we have

$\displaystyle \frac1{\sqrt{x^4+1}}<\frac1{x^2}.$

The problem is that ${1/x^2}$ has an asymptote at 0. However, we can split the integral as

$\displaystyle \int_0^\infty\frac{dx}{\sqrt{x^4+1}}=\int_0^1\frac{dx}{\sqrt{x^4+1}}+\int_1^\infty\frac{dx}{\sqrt{x^4+1}}.$

The first integral is finite, simply because it is the integral of a continuous function on some finite interval, there is nothing improper here. The second converges by comparison with the ${p}$-integral ${\displaystyle\int_1^\infty\frac{dx}{x^2}}$ with ${p=2>1.}$

It follows that the whole integral converges.

A slightly different approach is to use limit comparison rather than direct comparison:

$\displaystyle \lim_{x\rightarrow\infty}\frac{1/\sqrt{x^4+1}}{1/x^2}=\lim_{x\rightarrow\infty}\sqrt{\frac{x^4}{x^4+1}}=\sqrt1=1.$

It follows that ${\displaystyle\int_1^\infty\frac{dx}{\sqrt{x^4+1}}}$ converges because ${\displaystyle\int_1^\infty\frac{dx}{x^2}}$ does; and we have to treat

$\displaystyle \int_0^1\frac{dx}{\sqrt{x^4+1}}$

as above. We cannot just compare ${\displaystyle\int_0^\infty\frac{dx}{\sqrt{x^4+1}}}$ with ${\displaystyle\int_0^\infty\frac{dx}{x^2}}$ (which happens to diverge), because the limit comparison test requires that the functions that we compare are continuous, and ${1/x^2}$ is not continuous at 0.

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