An integral \(\displaystyle \int_a^b f(x) \,dx\) is called an improper integral if

1. \(a=-\infty\) or \(b=\infty\), i.e., the interval of integration is an infinite interval, or,
   
   
2. \(f\) has an infinite discontinuity at some point \(c\) in \([a,b]\), i.e., \[ \lim_{x\to c^+} f(x)=\pm \infty \text{ or }\lim_{x\to c^-} f(x)=\pm \infty. \]

Infinite Intervals:

1. If \(\displaystyle \int_a^t f(x) \,dx\) exists for all \(t\geq a\), then we define \(\displaystyle \int_a^{\infty} f(x) \,dx\) as the following limit when it exists: \[ \int_a^{\infty} f(x) \,dx=\lim_{t\to \infty} \int_a^t f(x) \,dx. \]
2. If \(\displaystyle \int_t^a f(x) \,dx\) exists for all \(t\leq a\), then we define \(\displaystyle \int_{-\infty}^a f(x) \,dx\) as the following limit when it exists: \[ \int_{-\infty}^a f(x) \,dx=\lim_{t\to -\infty} \int_t^a f(x) \,dx. \]
3. We define \(\displaystyle \int_{-\infty}^{\infty} f(x) \,dx\) as \[ \int_{-\infty}^{\infty} f(x) \,dx=\int_{-\infty}^a f(x) \,dx + \int_a^{\infty} f(x) \,dx, \] provided both \(\displaystyle \int_{-\infty}^a f(x) \,dx\) and \(\displaystyle \int_a^{\infty} f(x) \,dx\) exist.
   An improper integral is convergent if the corresponding limit exists. Otherwise it is divergent.

Example. Evaluate \(\displaystyle \int_0^{\infty} \frac{dx}{x^2+1}\) and \(\displaystyle \int_{-\infty}^{\infty} \frac{dx}{x^2+1}\).
Solution. \[\begin{align*} \int_0^{\infty} \frac{dx}{x^2+1} &=\lim_{t\to \infty} \int_0^t \frac{dx}{x^2+1}\\ &=\lim_{t\to \infty} \left.\tan^{-1}x \right|_0^t \\ &=\lim_{t\to \infty} \left( \tan^{-1}t - \tan^{-1}0 \right)\\ &=\lim_{t\to \infty} \tan^{-1}t\\ &=\frac{\pi}{2}. \end{align*}\] Similarly, \[\begin{align*} \int_{-\infty}^0 \frac{dx}{x^2+1} &=\lim_{t\to -\infty} \int_t^0 \frac{dx}{x^2+1}\\ &=\lim_{t\to -\infty} \left.\tan^{-1}x \right|_t^0 \\ &=\lim_{t\to -\infty} \left(-\tan^{-1}t \right)\\ &=\frac{\pi}{2}. \end{align*}\] Note that the preceding integral can also be obtained by observing the symmetry of the function \(f(x)=\displaystyle\frac{1}{x^2+1}\). Since both \(\displaystyle\int_{-\infty}^0 \frac{dx}{x^2+1}\) and \(\displaystyle\int_0^{\infty} \frac{dx}{x^2+1}\) are convergent, \[ \int_{-\infty}^{\infty} \frac{dx}{x^2+1} =\int_{-\infty}^0 \frac{dx}{x^2+1}+ \int_0^{\infty} \frac{dx}{x^2+1}=\frac{\pi}{2}+\frac{\pi}{2}=\pi. \]

Example. For \(a >0\), show that \(\displaystyle \int_a^{\infty} \frac{1}{x^p} \,dx=\frac{a^{1-p}}{p-1}\) if \(p >1\) and the integral is divergent if \(p\leq 1\).
Solution. First note that \[ \int \frac{1}{x^p}\,dx= \begin{cases} \frac{x^{1-p}}{1-p} & \text{if } p\neq 1 \\ \ln x & \text{if } p=1. \end{cases} \] If \( p=1\), then \[ \int_a^{\infty} \frac{1}{x} \,dx =\lim_{t\to \infty} \int_a^t \frac{1}{x}\,dx =\lim_{t\to \infty} \left.\ln x \right|_a^t =\lim_{t\to \infty} \left( \ln t - \ln a \right) =\infty. \] If \(p< 1\), then \[ \int_a^{\infty} \frac{1}{x^p} \,dx =\lim_{t\to \infty} \int_a^t \frac{1}{x^p}\,dx =\lim_{t\to \infty} \left.\frac{x^{1-p}}{1-p} \right|_a^t =\lim_{t\to \infty} \left( \frac{t^{1-p}}{1-p} - \frac{a^{1-p}}{1-p} \right) =\infty \;(\text{as } 1-p>0). \] Thus the integral is divergent if \(p\leq 1\). If \( p>1\), then \[ \int_a^{\infty} \frac{1}{x^p} \,dx =\lim_{t\to \infty} \int_a^t \frac{1}{x^p}\,dx =\lim_{t\to \infty} \left.\frac{x^{1-p}}{1-p} \right|_a^t =\lim_{t\to \infty} \left( \frac{t^{1-p}}{1-p} - \frac{a^{1-p}}{1-p} \right) =0- \frac{a^{1-p}}{1-p} \;(\text{as } 1-p<0). \]

Infinite Discontinuities:

1. If \(f\) is continuous on \([a,b)\) and \(f\) has an infinite discontinuity at \(b\), i.e., \(\displaystyle \lim_{x\to b^-} f(x)=\pm \infty\), then we define \(\displaystyle \int_a^b f(x) \,dx\) as follow: \[ \int_a^b f(x) \,dx= \lim_{t\to b^-} \int_a^t f(x) \,dx. \]
2. If \(f\) is continuous on \((a,b]\) and \(f\) has an infinite discontinuity at \(a\), i.e., \(\displaystyle \lim_{x\to a^+} f(x)=\pm \infty\), then we define \(\displaystyle \int_a^b f(x) \,dx\) as follow: \[ \int_a^b f(x) \,dx= \lim_{t\to a^+} \int_t^b f(x) \,dx. \]
3. If \(f\) has an infinite discontinuity at \(c\), \(a< c< b\), then we define \(\displaystyle \int_a^b f(x) \,dx\) as follow: \[ \int_a^b f(x) \,dx=\int_a^c f(x) \,dx+\int_c^b f(x) \,dx, \] provided both \(\displaystyle \int_a^c f(x) \,dx\) and \(\displaystyle \int_c^b f(x) \,dx\) exist.
   An improper integral is convergent if the corresponding limits exist. Otherwise it is divergent.

Example. Show that \(\displaystyle\int_0^1 \ln x \,dx=-1\).
Solution. First note that \(\ln\) is continuous on \((0,1]\) and \(\ln\) has an infinite discontinuity at \(0\) as \(\displaystyle \lim_{x\to 0^+} \ln x=- \infty\). \[\begin{align*} \int_0^1 \ln x \,dx &=\lim_{t\to 0^+} \int_t^1 \ln x \,dx\\ &=\lim_{t\to 0^+} \left. x\ln x-x \right|_t^1\\ &=\lim_{t\to 0^+} \left(-1-t\ln t+t \right)\\ &=-1-\lim_{t\to 0^+} t\ln t\\ &=-1-\lim_{t\to 0^+} \frac{\ln t}{\frac{1}{t}} \;\left(\frac{\infty}{\infty} \right)\\ &=-1-\lim_{t\to 0^+} \frac{\frac{1}{t}}{-\frac{1}{t^2}} \;\left(\text{by L'Hospital's Rule} \right)\\ &=-1+\lim_{t\to 0^+} t\\ &=-1. \end{align*}\]

Example. Show that \(\displaystyle\int_{-1}^1 \frac{dx}{x}\) is divergent.
Solution. First note that the fundamental theorem of calculus is not applicable here as the integrand is not defined at \(0\). Consequently, the following is incorrect: \[ \int_{-1}^1 \frac{dx}{x}=\left.\ln|x| \right\vert_{-1}^1=\ln|1|-\ln|-1|=0. \] Note that \(\frac{1}{x}\) is continuous on \([-1,0)\cup (0,1]\) and it has an infinite discontinuity at \(0\) as \(\displaystyle \lim_{x\to 0^+} \frac{1}{x}=\infty\). \[\begin{align*} \int_0^1 \frac{dx}{x} &=\lim_{t\to 0^+} \int_t^1 \frac{dx}{x}\\ &=\lim_{t\to 0^+} \left. \ln|x| \right|_t^1\\ &=\lim_{t\to 0^+} \left(0-\ln|t| \right)\\ &=\infty. \end{align*}\] Then \(\displaystyle\int_0^1 \frac{dx}{x}\) is divergent and consequently \(\displaystyle\int_{-1}^1 \frac{dx}{x}\) is divergent.

Comparison Test:
Suppose that \(f\) and \(g\) are continuous and \( 0\leq g(x)\leq f(x)\) for all \(x\geq a\).

1. If \(\displaystyle \int_a^{\infty} f(x) \,dx\) is convergent, then \(\displaystyle \int_a^{\infty} g(x) \,dx\) is also convergent.
   
   
2. If \(\displaystyle \int_a^{\infty} g(x) \,dx\) is divergent, then \(\displaystyle \int_a^{\infty} f(x) \,dx\) is also divergent.

Example. Use the Comparison Test to show that \(\displaystyle\int_1^{\infty} \frac{dx}{x^2+e^{-x}}\) is convergent.
Solution. For all \(x\geq 1\), \[ 0\leq \frac{1}{x^2+e^{-x}}\leq \frac{1}{x^2}. \] Now \(\displaystyle \int_1^{\infty} \frac{1}{x^2} \,dx\) is convergent by the integral p-test where \(p=2>1\). By the Comparison Test, \(\displaystyle\int_1^{\infty} \frac{dx}{x^2+e^{-x}}\) is convergent.
Now we will see in the following how the choice of a bigger function matters: for all \(x\geq 1\), \[ 0\leq \frac{1}{x^2+e^{-x}}\leq \frac{1}{e^{-x}}. \] Now \(\displaystyle \int_1^{\infty} \frac{1}{e^{-x}} \,dx=\int_1^{\infty} e^x \,dx\) is divergent. Then we cannot apply the Comparison Test.

Last edited