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.

Now we return to the problem:

\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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This entry was posted on Friday, November 13th, 2009 at 2:32 pm and is filed under 175: Calculus II. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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