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## Improper Integrals

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.

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