4.30 Improper Integral

1. \(\displaystyle\int_0^\infty (1+2x)e^{-x} dx\)

   Consider \(\displaystyle\int_0^t (1+2x)e^{-x} dx\)
   Let \(u(x) = 1+2x\), \(v'(x) = e^{-x}\), then \(u'(x)=2\), \(v(x) = -e^{-x}\)
   Then \(\displaystyle\int_0^t (1+2x)e^{-x} dx = -(1+2x)e^{-x} \Big|_0^t - \displaystyle\int_0^t -2e^{-x} dx\)
   Then \(\displaystyle\int_0^t (1+2x)e^{-x} dx = -(1+2x)e^{-x} \Big|_0^t + \displaystyle\int_0^t 2e^{-x} dx\)
   ​\(=-(1+2t)e^{-t}+1+(-2e^{-x}\Big|_{0}^{t})\)​
   ​\(= -2te^{-t} - 3e^{-t} + 3\)

   Then \(\lim_{t \to \infty} \displaystyle\int_0^t (1+2x)e^{-x} dx = \lim_{t \to \infty} (-2te^{-t} - 3e^{-t} + 3)\) \(=\lim_{t\to\infty}\left(\frac{-2t}{e^{t}}\right)+3=\lim_{t\to\infty}\frac{-2}{e^{t}} +3=3\)

2. \(\displaystyle\int_{-\infty}^{\infty} \frac{6x^3}{(x^4+1)^2} dx\)

   Consider \(\displaystyle\int_{-\infty}^{\infty} \frac{6x^3}{(x^4+1)^2} dx = \displaystyle\int_{-\infty}^{0} \frac{6x^3}{(x^4+1)^2} dx + \displaystyle\int_{0}^{\infty} \frac{6x^3}{(x^4+1)^2} dx\)

   For the \(\displaystyle\int_{0}^{t} \frac{6x^3}{(x^4+1)^2} dx\), Let \(u = x^4+1\), then \(du = 4x^3 dx\).
   ​\(\displaystyle\int_{0}^{t}\frac{6x^{3}}{(x^{4}+1)^{2}}dx = \displaystyle\int_{1}^{t^4+1} \frac{6}{u^{2}}\frac{du}{4}= \displaystyle\int_{1}^{t^4+1}\frac{3}{2u^{2}}du=-\frac{3}{2(t^{4}+1)}+\frac{3}{2}\)

   Then \(\displaystyle\int^{0}_{-\infty}\frac{6x^{3}}{(x^{4}+1)^{2}}dx =- \lim_{t \to \infty}\left(-\frac{3}{2(t^{4}+1)}+ \frac{3}{2}\right) = -\frac{3}{2}\)

   Then \(\displaystyle\int_{0}^{\infty} \frac{6x^3}{(x^4+1)^2} dx = \lim_{t \to \infty} \left(-\frac{3}{2(t^4+1)} + \frac{3}{2}\right) = 0 + \frac{3}{2} = \frac{3}{2}\)

   Thus \(\displaystyle\int_{-\infty}^{\infty} \frac{6x^3}{(x^4+1)^2} dx = \displaystyle\int_{-\infty}^{0} \frac{6x^3}{(x^4+1)^2} dx + \displaystyle\int_{0}^{\infty} \frac{6x^3}{(x^4+1)^2} dx = -\frac{3}{2} + \frac{3}{2} = 0\)​

3. \(\displaystyle\int_1^4 \frac{1}{x^2+x-6} dx = \displaystyle\int_1^4 \frac{1}{(x+3)(x-2)} dx = \displaystyle\int_1^2 \frac{1}{(x+3)(x-2)} dx + \displaystyle\int_2^4 \frac{1}{(x+3)(x-2)} dx\)

   Then consider \(\displaystyle\int \frac{1}{(x+3)(x-2)} dx\), since \(\frac{1}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}=\frac{Ax-2A+Bx+3B}{(x+3)(x-2)}\)
   Then \(A+B=0\), \(-2A+3B=1\). \(A=-B\). \(-2(-B)+3B=1 \implies 2B+3B=1 \implies 5B=1 \implies B=\frac{1}{5}\), \(A=-\frac{1}{5}\).
   Then \(\displaystyle\int \frac{1}{(x+3)(x-2)} dx = \displaystyle\int -\frac{1}{5(x+3)} dx + \displaystyle\int \frac{1}{5(x-2)} dx = -\frac{1}{5}\ln|x+3| + \frac{1}{5}\ln|x-2| = \frac{1}{5}(\ln|x-2| - \ln|x+3|)\)

   Then \(\displaystyle\int_1^2 \frac{1}{(x+3)(x-2)} dx = \lim_{t \to 2^-} \left[\frac{1}{5}(\ln|x-2| - \ln|x+3|)\right]_1^t\)​
   ​\(= \lim_{t \to 2^-}\frac{1}{5}(\ln|t-2|) - \lim_{t \to 2^-}\frac{1}{5}(\ln|t+3|) - \frac{1}{5}(\ln(1) - \ln(4))\)
   ​\(= \frac{1}{5} (-\infty) - \frac{1}{5}\ln(5) - \frac{1}{5}(0 - \ln(4))\)
   ​\(= -\infty\)

4. \(\displaystyle\int_{-\infty}^0 \frac{e^{1/x}}{x^2} dx = \displaystyle\int_{-\infty}^{-1} \frac{e^{1/x}}{x^2} dx + \displaystyle\int_{-1}^0 \frac{e^{1/x}}{x^2}\)

   For \(\displaystyle\int \frac{e^{1/x}}{x^2} dx\), let \(t = e^{1/x}\), then \(dt = e^{1/x} \cdot (-\frac{1}{x^2}) dx = -\frac{e^{1/x}}{x^2} dx\).
   Then \(\displaystyle\int \frac{e^{1/x}}{x^2} dx = \displaystyle\int -dt = -t + C = -e^{1/x} + C\).

   Thus \(\displaystyle\int_{-\infty}^{-1} \frac{e^{1/x}}{x^2} dx = \lim_{n \to -\infty} \displaystyle\int_n^{-1} \frac{e^{1/x}}{x^2} dx = \lim_{n \to -\infty} [-e^{1/x}]_n^{-1} = \lim_{n \to -\infty} (-e^{-1} - (-e^{1/n})) = \lim_{n \to -\infty} (-e^{-1} + e^{1/n}) = -e^{-1} + 1\).
   ​\(\displaystyle\int_{-1}^0 \frac{e^{1/x}}{x^2} dx = \lim_{n \to 0^-} \displaystyle\int_{-1}^n \frac{e^{1/x}}{x^2} dx = \lim_{n \to 0^-} [-e^{1/x}]_{-1}^n = \lim_{n \to 0^-} (-e^{1/n} - (-e^{-1})) = \lim_{n \to 0^-} (-e^{1/n} + e^{-1}) = 0 + e^{-1} = e^{-1}\).

   Thus result \(= (-e^{-1} + 1) + e^{-1} = 1\).

5. \(\displaystyle\int_0^\infty \frac{\arctan(x)}{1+x^2} dx = \lim_{t \to \infty} \displaystyle\int_0^t \frac{\arctan(x)}{1+x^2} dx\)

   Let \(\arctan(x) = u\), then \(\frac{1}{1+x^2} dx = du\).
   Then \({\displaystyle\int_0^{t}\frac{\arctan(x)}{1+x^{2}}dx={\displaystyle\int_0^{\arctan(t)}udu=\frac{1}{2}(\arctan(t))^2}}\)

   Then \(\displaystyle\int_0^\infty \frac{\arctan(x)}{1+x^2} dx = \lim_{t \to \infty} \frac{1}{2} (\arctan(t))^2 = \frac{1}{2} \left(\frac{\pi}{2}\right)^2 = \frac{1}{2} \frac{\pi^2}{4} = \frac{\pi^2}{8}\)

   How to calculate \(\arctan(x)'\): let \(\arctan(x) = y\), then \(\tan(y) = x\).
   Differentiate with respect to \(x\): \(\sec^2(y) \cdot y' = 1\).
   ​\(y' = \frac{1}{\sec^2(y)} = \frac{1}{1+\tan^2(y)} = \frac{1}{1+x^2}\).

6. \(\displaystyle\int_1^\infty \frac{\arctan(x)}{x^2+\ln(x)} dx = \lim_{t \to \infty} \displaystyle\int_1^t \frac{\arctan(x)}{x^2+\ln(x)} dx\)

   Consider \(\displaystyle\int_1^t \frac{\arctan(x)}{x^2+\ln(x)} dx\), using comparison test.
   We have \(\frac{\arctan(x)}{x^2+\ln(x)} \le \frac{\pi/2}{x^2+\ln(x)} \le \frac{\pi/2}{x^2}\).
   Since we know \(\displaystyle\int_1^\infty \frac{1}{x^2} dx\) is convergent, then the integral \(\displaystyle\int_1^\infty \frac{\pi}{2} \frac{1}{x^2} dx\) is also convergent.
   Then the integral \(\displaystyle\int_1^\infty \frac{\arctan(x)}{x^2+\ln(x)} dx\) is convergent.

7. \(\displaystyle\int_0^\infty \arctan(\frac{1}{x}) dx = \int_0^1 \arctan(\frac{1}{x}) dx + \int_1^\infty \arctan(\frac{1}{x}) dx\)

   Consider \(\displaystyle\int \arctan(\frac{1}{x}) dx\), let \(\arctan(\frac{1}{x}) = u(x)\), \(1 = v'(x)\)
   Then \(u'(x) = \frac{1}{1+(\frac{1}{x})^2} \times (-\frac{1}{x^2}) = \frac{x^2}{x^2+1} \times (-\frac{1}{x^2}) = -\frac{1}{1+x^2}\). \(v(x) = x\)
   Then \(\displaystyle\int \arctan(\frac{1}{x}) dx = \arctan(\frac{1}{x}) \cdot x - \int x \cdot (-\frac{1}{1+x^{2}} ) dx = \arctan(\frac{1}{x}) \cdot x + \int \frac{x}{1+x^{2}}dx = \arctan(\frac{1}{x} ) \cdot x + \frac{1}{2}\ln(x^{2}+1)\)

   Then \(\displaystyle\int_0^1 \arctan(\frac{1}{x}) dx = \frac{\pi}{4} - \lim_{x \to 0^+} \arctan(\frac{1}{x}) x + \frac{1}{2}\ln 2\) \(= \frac{\pi}{4} + \frac{1}{2}\ln 2\)

   \(\displaystyle\int_1^\infty \arctan(\frac{1}{x}) dx = \lim_{x \to \infty} \arctan(\frac{1}{x}) x - \frac{\pi}{4} + \lim_{x \to \infty} \frac{1}{2}\ln(x^2+1) - \frac{\pi}{4}\)
   \(= \lim_{x \to \infty} \frac{\arctan(\frac{1}{x})}{1/x} - \frac{\pi}{4} + \lim_{x \to \infty} \frac{1}{2}\ln(x^2+1)\)
   \(= \lim_{x \to \infty} \frac{\frac{-1}{1+(1/x)^2} \cdot (-\frac{1}{x^2})}{-\frac{1}{x^2}} - \frac{\pi}{4} + \lim_{x \to \infty} \frac{1}{2}\ln(x^2+1)\)
   \(= 1 - \frac{\pi}{4} + \infty - \frac{\pi}{4}\)
   \(= 1 - \frac{\pi}{2} + \infty\)

   Thus the integral doesn't exist.

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