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san4es73 [151]
3 years ago
9

Determine whether the integral converges.

Mathematics
1 answer:
Kryger [21]3 years ago
7 0
You have one mistake which occurs when you integrate \dfrac1{1-p^2}. The antiderivative of this is not in terms of \tan^{-1}p. Instead, letting p=\sin r (or \cos r, if you want to bother with more signs) gives \mathrm dp=\cos r\,\mathrm dr, making the indefinite integral equality

\displaystyle-\frac12\int\frac{\mathrm dp}{1-p^2}=-\frac12\int\frac{\cos r}{1-\sin^2r}\,\mathrm dr=-\frac12\int\sec r\,\mathrm dr=\ln|\sec r+\tan r|+C

and then compute the definite integral from there.

-\dfrac12\ln|\sec r+\tan r|\stackrel{r=\sin^{-1}p}=-\dfrac12\ln\left|\dfrac{1+p}{\sqrt{1-p^2}}=\ln\left|\sqrt{\dfrac{1+p}{1-p}}\right|
\stackrel{p=u/2}=-\dfrac12\ln\left|\sqrt{\dfrac{1+\frac u2}{1-\frac u2}}\right|=-\dfrac12\ln\left|\sqrt{\dfrac{2+u}{2-u}}\right|
\stackrel{u=x+1}=-\dfrac12\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|
\implies-\dfrac12\displaystyle\lim_{t\to\infty}\ln\left|\sqrt{\dfrac{3+x}{1-x}}\right|\bigg|_{x=2}^{x=t}=-\frac12\left(\ln|-1|-\ln\left|\sqrt{\frac5{-1}}\right|\right)=\dfrac{\ln\sqrt5}2=\dfrac{\ln5}4

Or, starting from the beginning, you could also have found it slightly more convenient to combine the substitutions in one fell swoop by letting x+1=2\sec y. Then \mathrm dx=2\sec y\tan y\,\mathrm dy, and the integral becomes

\displaystyle\int_2^\infty\frac{\mathrm dx}{(x+1)^2-4}=\int_{\sec^{-1}(3/2)}^{\pi/2}\frac{2\sec y\tan y}{4\sec^2y-4}\,\mathrm dy
\displaystyle=\frac12\int_{\sec^{-1}(3/2)}^{\pi/2}\csc y\,\mathrm dy
\displaystyle=-\frac12\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2}}^{y=\pi/2}
\displaystyle=-\frac12\lim_{t\to\pi/2^-}\ln|\csc y+\cot y|\bigg|_{y=\sec^{-1}(3/2)}^{y=t}
\displaystyle=-\frac12\left(\lim_{t\to\pi/2^-}\ln|\csc t+\cot t|-\ln\frac5{\sqrt5}\right)
=\dfrac{\ln\sqrt5}2-\dfrac{\ln|1|}2
=\dfrac{\ln5}4

Another way to do this is to notice that the integrand's denominator can be factorized.

x^2+2x-3=(x+3)(x-1)

So,

\dfrac1{x^2+2x-3}=\dfrac1{(x+3)(x-1)}=\dfrac14\left(\dfrac1{x-1}-\dfrac1{x+3}\right)

There are no discontinuities to worry about since you're integrate over [2,\infty), so you can proceed with integrating straightaway.

\displaystyle\int_2^\infty\frac{\mathrm dx}{x^2+2x-3}=\frac14\lim_{t\to\infty}\int_2^t\left(\frac1{x-1}-\frac1{x+3}\right)\,\mathrm dx
=\displaystyle\frac14\lim_{t\to\infty}(\ln|x-1|-\ln|x+3|)\bigg|_{x=2}^{x=t}
=\displaystyle\frac14\lim_{t\to\infty}\ln\left|\frac{x-1}{x+3}\right|\bigg|_{x=2}^{x=t}
=\displaystyle\frac14\left(\lim_{t\to\infty}\ln\left|\frac{t-1}{t+3}\right|-\ln\frac15\right)
=\displaystyle\frac14\left(\ln1-\ln\frac15\right)
=-\dfrac14\ln\dfrac15=\dfrac{\ln5}4

Just goes to show there's often more than one way to skin a cat...
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