# Type 1: infinite intervals.

• Let $$f(x)$$ be a function, which has at most finite discontinuities on the interval $$[a,b]$$. Then the definite integral $\int_a^bf(x)\,\mathrm dx$ exists, and its value is the signed area between the graph $$y=f(x)$$ and the $$x$$-axis $$y=0$$.
• Now we'll see how to define and calculate areas over infinite intervals.

# Type 1: infinite intervals.

• Definition.
• If $$\int_a^tf(x)\,\mathrm dx$$ exists for all $$t\ge a$$, then $\int_a^\infty f(x)\,\mathrm dx=\lim_{t\to\infty}\int_a^tf(x)\,\mathrm dx,$ provided this limit exists (as a finite number).
• If $$\int_t^bf(x)\,\mathrm dx$$ exists for all $$t\le b$$, then $\int_{-\infty}^b f(x)\,\mathrm dx=\lim_{t\to-\infty}\int_t^bf(x)\,\mathrm dx,$ provided this limit exists (as a finite number).
• The integrals $$\int_a^\infty f(x)\,\mathrm dx$$ and $$\int_{-\infty}^bf(x)\,\mathrm dx$$ are called improper integrals. If the corresponding limit exists, then we say that the integral is convergent. Otherwise, we say it's divergent.
• If both $$\int_a^\infty f(x)\,\mathrm dx$$ and $$\int_{-\infty}^bf(x)\,\mathrm dx$$ are convergent, then we let $\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_{-\infty}^af(x)\,\mathrm dx+\int^\infty_af(x)\,\mathrm dx.$ It can be proven that the value is independent of the choice of $$a$$ (7.8.76).

# Type 1: infinite intervals.

• $$\int_1^\infty\frac{1}{x^3}\,\mathrm dx$$.
• For $$t\ge1$$, we have $\int_1^t\frac{1}{x^3}\,\mathrm dx=\left[-\frac{1}{2x^2}\right]_1^t=\frac12-\frac{1}{2t^2}.$
• Therefore, we get $\int_1^\infty\frac{1}{x^3}\,\mathrm dx=\lim_{t\to\infty}\frac{1}{x^3}\,\mathrm dx=\lim_{t\to\infty}\left(\frac12-\frac{1}{2t^2}\right)=\frac12,$ that is the improper integral is convergent.
• $$\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx$$.
• For $$t\ge1$$, we have $\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx=\left[2\sqrt x\right]_1^t=2t-2.$
• Therefore, we get $\int_1^\infty\frac{1}{\sqrt x}\,\mathrm dx=\lim_{t\to\infty}\int_1^t\frac{1}{\sqrt x}\,\mathrm dx=\lim_{t\to\infty}(2t-2)=\infty,$ that is the improper integral is divergent.

# Type 1: infinite intervals.

• For which $$p$$ is the integral $$\int_1^\infty x^p\,\mathrm dx$$ convergent?
• If $$p\ne-1$$, then we have $\int_1^\infty x^p\,\mathrm dx=\lim_{t\to\infty}\left[\frac{x^{p+1}}{p+1}\right]_1^t=\lim_{t\to\infty}\frac{t^{p+1}-1}{p+1}.$ This is convergent precisely when $$p<-1$$.
• If $$p=-1$$, then we have $\int_1^\infty x^{-1}\,\mathrm dx=\lim_{t\to\infty}\left[\ln x\right]_1^t=\lim_{t\to\infty}\ln t=\infty.$
• Therefore, the improper integral $$\int_1^\infty x^p\,\mathrm dx$$ is convergent precisely when $$p<1$$.

# Type 1: infinite intervals.

• $$\int_{-\infty}^02^x\,\mathrm dx$$.
• We have $\int_{-\infty}^02^x\,\mathrm dx=\frac{1}{\ln2}\lim_{t\to-\infty}\left(\left[x2^x\right]_t^0-\int_t^02^x\,\mathrm dx\right)=\lim_{t\to-\infty}\left(\frac{t2^t}{\ln 2}-\frac{2^t-1}{(\ln2)^2}\right)=\frac{1}{(\ln2)^2},$ therefore the integral is convergent.
• $$\int_{-\infty}^\infty\frac{1}{\sqrt{1+x^2}}\,\mathrm dx$$.
• We have $\int\frac{1}{\sqrt{1+x^2}}\,\mathrm dx\stackrel{x=\tan\theta}{=}\int\frac{\sec^2\theta}{\sec\theta}\,\mathrm d\theta=\ln|\tan\theta+\sec\theta|+C=\ln|x+\sqrt{1+x^2}|+C,$ which gives $\int_0^\infty\frac{1}{\sqrt{1+x^2}}\,\mathrm dx=\lim_{t\to\infty}\left[\ln|x+\sqrt{1+x^2}|\right]_0^t=\lim_{t\to\infty}\left(\ln(t+\sqrt{1+t^2})\right)=\infty,$ therefore the improper integral is divergent.
• Exercises. 7.8: 3, 5, 11, 15, 21, 25

# Type 2: Discontinuous integrals

• Another way to get improper integrals is to integrate functions with infinite discontinuities.
• Definition.
• If $$f$$ is continuous on $$[a,b)$$, and it has an infinite discontinuity at $$b$$, then $\int_a^bf(x)\,\mathrm dx=\lim_{t\to b}\int_a^tf(x)\,\mathrm dx,$ provided that the limit exists (as a finite number).
• If $$f$$ is continuous on $$(a,b]$$, and it has an infinite discontinuity at $$a$$, then $\int_a^bf(x)|,\mathrm dx=\lim_{t\to a}\int_t^bf(x)\,\mathrm dx,$ provided that the limit exists (as a finite number).
• In these cases, the integral $$\int_a^bf(x)\,\mathrm dx$$ is called an improper integral. If the limits exists, then it is convergent, and if the limit does not exist, then it is divergent.
• If $$f$$ has an infinite discontinuity at $$c$$ for $$a<c<b$$, and the improper integrals $$\int_a^cf(x),\mathrm dx$$ and $$\int_c^bf(x)\,\mathrm dx$$ are convergent, then we let $\int_a^bf(x)\,\mathrm dx=\int_a^cf(x)\,\mathrm dx+\int_c^bf(x)\,\mathrm dx.$

# Type 2: Discontinuous integrals

• $$\int_{-1}^3\frac{1}{\sqrt{x+1}}\,\mathrm dx$$.
• Note that $$f(x)=\frac{1}{\sqrt{x+1}}$$ has an infinite discontinuity at $$x=-3$$.
• For $$-1<t\le 3$$, we have $\int_t^3\frac{1}{\sqrt{x+1}}\,\mathrm dx=\left[2\sqrt{x+1}\right]_t^3=4-2\sqrt{t+1}.$
• Therefore, we get $\int_{-1}^3\frac{1}{\sqrt{x+1}}=\lim_{t\to-1}(4-2\sqrt{t+1})=4,$ which shows that the improper integral is convergent.
• $$\int_0^{\pi/2}\tan x\,\mathrm dx$$.
• Note that $$f(x)=\tan x$$ has an infinite discontinuity at $$x=\frac{\pi}{2}$$.
• We have $\int_0^{\pi/2}\tan x\,\mathrm dx=\lim_{t\to\pi/2}\left[-\ln|\cos x|\right]_0^t=\lim_{t\to\pi/2}(-\ln(\cos t))=\infty,$ therefore the integral is divergent.

# Type 2: Discontinuous integrals

• $$\int_{-2}^2\frac{1}{x^2-1}\,\mathrm dx$$.
• Note that $$f(x)=\frac{1}{x^2-1}$$ has infinite discontinuities at $$x=-1,1$$.
• To get an antiderivative, we can use partial fractions. To solve $1=A(x+1)+B(x-1),$ we need $$A=\frac12$$ and $$B=-\frac12$$.
• Therefore, we get $\int\frac{1}{x^2-1}\,\mathrm dx=\frac{\ln|x-1|-\ln|x+1|}{2}+C.$
• This in turn gives $\int_{-2}^{-1}\frac{1}{x^2-1}\,\mathrm dx=\frac12\lim_{t\to-1}\left[\ln|x-1|-\ln|x+1|\right]_{-2}^t=\frac12\lim_{t\to-1}(\ln|t-1|-\ln|t+1|)=-\infty,$ therefore the integral is divergent.
• WARNING. Note that if we had not noticed the infinite discontinuities, then we would have gotten the wrong answer $\int_{-2}^2\frac{1}{x^2-1}\,\mathrm dx=\frac12\left[\ln|x-1|-\ln|x+1|\right]_{-2}^2=-\ln3.$
• Exercises. 7.8: 29, 35, 39

# A comparison test for improper integrals

• Theorem. Suppose that $$f(x)$$ and $$g(x)$$ are continuous functions with $$f(x)\ge g(x)\ge0$$ for $$x\ge a$$. Then we have the following.
• If $$\int_a^\infty f(x)\,\mathrm dx$$ is convergent, then so is $$\int_a^\infty g(x)\,\mathrm dx$$.
• If $$\int_a^\infty g(x)\,\mathrm dx$$ is divergent, then so is $$\int_a^\infty f(x)\,\mathrm dx$$.
• $$\int_0^\infty e^{-x^2}\,\mathrm dx$$.
• We have mentioned that the antiderivative of $$e^{-x^2}$$ is not an elementary function.
• We can write $\int_0^\infty e^{-x^2}\,\mathrm dx=\int_0^1e^{-x^2}\,\mathrm dx+\int_1^\infty e^{-x^2}\,\mathrm dx.$
• If $$x\ge1$$, then $$-x^2\le-x$$, and therefore $$e^{-x^2}\le e^{-x}$$.
• Then as we have $\int_1^\infty e^{-x}\,\mathrm dx=\lim_{t\to\infty}(e^{-1}-e^{-t})=e^{-1},$ the comparison theorem shows that $$\int_1^\infty e^{-x^2}\,\mathrm dx$$ is convergent, and thus so is $$\int_0^\infty e^{-x^2}\,\mathrm dx$$.
• Exercises. 7.8: 49, 51, 53.
• More exercises. 7.8: 55, 63, 73, 79, 81