When we defined the definite integral $\int_a^b f(x)dx$, it was assumed that the limits $a$ and $b$ of the integral are finite and the integrand $f(x)$ is continuous on the closed interval $[a,b]$. Even if these assumptions are not satisfied, we can still consider a notion of integral extended from the definite integral (the Riemann integral). This extended integral is called the improper integral.
Infinite Limits
A definite integral, in which one or both limits of integration are infinite, is defined by the following: \begin{align*}\int_a^\infty f(x)dx&=\lim_{t\to\infty}\int_a^tf(x)dx\\\int_{-\infty}^b f(x)dx&=\lim_{t\to -\infty}\int_t^bf(x)dx\end{align*} The improper integrals are said to be convergent if the corresponding limit exists. Otherwise, divergent.
If both $\int_{-\infty}^a f(x)dx$ and $\int_a^\infty f(x)dx$ are convergent, we define $$\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^a f(x)dx+\int_a^\infty f(x)dx$$
$\int_{-\infty}^\infty f(x)dx$ can be also defined by the double limit $$\int_{-\infty}^\infty f(x)dx=\lim_{b\to\infty\\a\to -\infty}\int_a^b f(x)dx$$
$\int_{-\infty}^\infty\frac{2x}{1+x^2}dx$ is divergent as $\int_{-\infty}^0\frac{2x}{1+x^2}dx=-\infty$ and $\int_0^\infty\frac{2x}{1+x^2}dx=\infty$. On the other hand, $$\lim_{a\to\infty}\int_{-a}^a\frac{2x}{1+x^2}dx=0$$ This is called the Cauchy principal value of the integral $\int_{-\infty}^\infty\frac{2x}{1+x^2}dx$ and is denoted by $$\mathrm{p.v.}\int_{-\infty}^\infty\frac{2x}{1+x^2}dx$$ The Cauchy principal value is a method of assigning values to certain ill-defined improper integrals. We will not, however, be considering the Cauchy principal value here.
Discontinuous Integrand
If $f(x)$ is continuous for all values of $x$ in the domain $a\leq x\leq b$ except $x=b$ or $x=a$, $\int_a^b f(x)dx$ is defined by \begin{equation}\label{eq:impropint}\int_a^b f(x)dx=\lim_{t\to b-}\int_a^t f(x)dx\end{equation} or \begin{equation}\label{eq:impropint2}\int_a^b f(x)dx=\lim_{t\to a+}\int_t^b f(x)dx\end{equation} provided the corresponding limit exists.
When $f(x)$ is continuous for all values of $x$ in the domain $a\leq x\leq b$ except $x=c$ (where $a< c <b$), $\int_a^b f(x)dx$ is defined by $$\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx$$ where the integrals in the RHS are evaluated in accordance with \eqref{eq:impropint} and \eqref{eq:impropint2}, respectively.
Example. Consider $\int_{-1}^1\frac{dx}{x^2}$.
The graph of y=1/x^2 on [-1,1]
Solution. Since the integrand is discontinous at $x=0$, we write the integral in two parts as $$\int_{-1}^1\frac{dx}{x^2}=\int_{-1}^0\frac{dx}{x^2}+\int_0^1\frac{dx}{x^2}=\infty+\infty=\infty$$
Remarks. If one mindlessly evaluates the integral as an ordinary definite integral, we obtain $$\int_{-1}^1\frac{dx}{x^2}=\left[-\frac{1}{x}\right]_{-1}^1=-2$$ However, this is nonsense because the integrand is always positive.
Let us consider the integral $\int\frac{5x-3}{x^2-2x-3}dx$. $\frac{d}{dx}(x^2-2x-3)=2x-2$ so substitution is not an option. Noting $x^2-2x-3=(x+1)(x-3)$, let us assume instead that $$\frac{5x-3}{x^2-2x-3}=\frac{A}{x+1}+\frac{B}{x-3}$$ Then \begin{align*}5x-3&=A(x-3)+B(x+1)\\&=(A+B)x-3A+B\end{align*} Hence we obtain a system of linear equations $$\left\{\begin{aligned}A+B&=5\\-3A+B&=-3\end{aligned}\right.$$ Solving this system simultaneously we find $A=2$ and $B=3$.
Alternation: Let’s begin with $5x-3=A(x-3)+B(x+1)$. For $x=3$, we get $12=4B$ so $B=3$. For $x=-1$, we get $-8=-4A$ so $A=2$. This method certainly has a computational advantage when it works over getting a system of linear equations and solving it.
Now \begin{align*}\int\frac{5x-3}{x^2-2x-3}dx&=2\int\frac{dx}{x+1}+3\int\frac{dx}{x-3}\\&=2\ln|x+1|+3\ln|x-3|+C\end{align*}
General method of writing a rational function $\frac{f(x)}{g(x)}$ as a sum of partial fractions
Let $x-r$ be a linear factor of $g(x)$. Suppose that $(x-r)^m$ is the highest power of $x-r$ that divides $g(x)$ i.e. $x=r$ is a zero of $g(x)$ with multiplicity $m$. Then to this factor, assign the sum of the $m$ partial fractions $$\frac{A_1}{x-r}+\frac{A_2}{(x-r)^2}+\cdots+\frac{A_m}{(x-r)^m}$$
Let $x^2+px+q$ be a quadratic factor of $g(x)$ that cannot be factored further into linear factors with real coefficients. Suppose that $(x^2+px+q)^n$ is the highest power of this factor that divides $g(x)$. Then to this factor assign the sum of the $n$ partial fractions $$\frac{B_1x+C_1}{x^2+px+q}+\frac{B_2x+C_2}{(x^2+px+q)^2}+\cdots+\frac{B_nx+C_n}{(x^2+px+q)^n}$$
Solution. Let $$\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3}$$ Then $$A(x+1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)$$ For $x=1$, $8A=6$ so $A=\frac{3}{4}$. For $x=-1$, $-4B=-2$ so $B=\frac{1}{2}$. For $x=-3$, $8C=-2$ so $C=-\frac{1}{4}$. Hence, \begin{align*}\int\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}dx&=\frac{3}{4}\int\frac{dx}{x-1}+\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{4}\int\frac{dx}{x+3}\\&=\frac{3}{4}\ln|x-1|+\frac{1}{2}\ln|x+1|-\frac{1}{4}\ln|x+3|+C\end{align*}
Example. Evaluate $\int\frac{6x+7}{(x+2)^2}dx$.
Solution. Let $\frac{6x+7}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$. Then $A=6$ and $B=-5$. Hence, \begin{align*}\int\frac{6x+7}{(x+2)^2}dx\\&=6\int\frac{dx}{x+2}-5\int\frac{dx}{(x+2)^2}\\&=6\ln|x+2|+\frac{5}{x+2}+C\end{align*}
Example. [Integrating an Improper Fraction] Evaluate $\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx$.
Solution. By long division, divide $2x^3-4x^2-x-3$ by $x^2-2x-3$ to obtain quotient $2x$ and remainder $5x-3$. Thus $2x^3-4x^2-x-3=(x^2-2x-3)\cdot 2x+5x-3$ and $$\frac{2x^3-4x^2-x-3}{x^2-2x-3}=2x+\frac{5x-3}{x^2-2x-3}$$ $\frac{5x-3}{x^2-2x-3}$ is a proper fraction so we can apply the above method to write it as $$\frac{5x-3}{x^2-2x-3}=\frac{3}{x-3}+\frac{2}{x+1}$$ Hence, \begin{align*}\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx&=\int 2xdx+3\int\frac{dx}{x-3}+2\int\frac{dx}{x+1}\\&=x^2+3\ln|x-3|+2\ln|x+1|+C\end{align*}
Remark. Instead of writing $\frac{5x-3}{x^2-2x-3}$ as $\frac{A}{x-3}+\frac{B}{x+1}$, let $$5x-3=A(x+1)+B(x-3)$$ $\frac{5x-3}{x+1}=A+\frac{B(x-3)}{x+1}$ and if $x=3$, $A=3$. $\frac{5x-3}{x-3}=\frac{A(x+1)}{x-3}+B$ and if $x=-1$, $B=2$. This is called Heaviside’s method and is easier to determine coefficients than the standard method we discussed above. However it can be useful only when the denominator has all linear factors.
Solution. Let $$\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$$ Then we obtain \begin{align*}-2x+4&=(Ax+B)(x-1)^2+C(x-1)(x^2+1)+D(x^2+1)\\&=(A+C)x^3+(-2A+B-C+D)x^2+(A-2B+C)x+(B-C+D)\end{align*} and hence the equations $A+C=0$, $-2A+B-C+D=0$, $A-2B+C=-2$, and $B-C+D=4$. Solve these equations simultaneously to obtain $A=2$, $B=1$, $C=-2$, and $D=1$. Therefore, \begin{align*}\int\frac{-2x+4}{(x^2+1)(x-1)^2}dx&=\int\frac{2x+1}{x^2+1}dx-2\int\frac{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx-2\int{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\ln(x^2+1)+\tan^{-1}x-2\ln|x-1|-\frac{1}{x-1}+C\end{align*}
Example. Evaluate $\int\frac{dx}{x(x^2+1)^2}$.
Solution. Let $$\frac{1}{x(x^2+1)^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$ Then we have \begin{align*}1&=A(x^2+1)^2+(Bx+C)x(x^2+1)+(Dx+E)x\\&=(A+B)x^4+Cx^3+(2A+B+D)x^2+(C+E)x+A\end{align*} and comparing the coefficients we obtain the equations $A+B=0$, $C=0$, $2A+B+D=0$, $C+E=0$, and $A=1$. Solve these equations simultaneously to obtain $A=1$, $B=-1$, $C=0$, $D=-1$, and $E=0$. Therefore, \begin{align*}\int\frac{dx}{x(x^2+1)^2}&=\int\frac{dx}{x}-\int\frac{x}{x^2+1}dx-\int\frac{x}{(x^2+1)^2}dx\\&=\ln|x|-\frac{1}{2}\ln(x^2+1)+\frac{1}{2(x^2+1)}+C\end{align*}
Let us consider forced harmonic oscillation without damping $$m\ddot{X}(t)+kX(t)=F(t)$$ with initial conditions $X(0)=x_0$ and $\dot{X}(0)=v_0$. The transformed equation is $$m[s^2x(s)-sx_0-v_0]+kx(s)=f(s)$$ from which we obtain \begin{equation}\begin{aligned} x(s)&=\frac{sx_0+v_0}{s^2+\omega_0^2}+\frac{1}{m}\frac{f(s)}{s^2+\omega_0^2}\\&=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{f(s)}{m\omega_0}\frac{\omega_0}{s^2+\omega_0^2} \end{aligned}\label{eq:laplace18}\end{equation} where $\omega_0=\sqrt{\frac{k}{m}}$. By taking the inverse transform of \eqref{eq:laplace18} we find the solution $X(t)$ as \begin{equation} \begin{aligned} X(s)&=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{1}{m\omega_0}F(t)\ast\cos\omega_0t\\ &=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{1}{m\omega_0}\int_0^t\sin\omega_0(t-\tau)F(\tau)d\tau \end{aligned}\label{eq:laplace19}\end{equation}
Resonance
Let $F(t)$ be the sinusoidal force $F(t)=F_0\sin\omega t$ where $F_0$ and $\omega$ are positive constants. Then
$$f(s)=\mathcal{L}{F(t)}=F_0\frac{\omega}{s^2+\omega^2}$$
and \eqref{eq:laplace18} can be written
$$x(s)=\frac{x_0s+v_0}{s^2+\omega_0^2}+\frac{F_0}{m}\frac{\omega}{(s^2+\omega_0^2)(s^2+\omega^2)}$$
If $\omega\ne\omega_0$, $\frac{1}{(s^2+\omega_0^2)(s^2+\omega^2)}=\frac{1}{\omega^2-\omega_0^2}\left[\frac{1}{s^2+\omega_0^2}-\frac{1}{s^2+\omega^2}\right]$ and
$$x(s)=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{F_0\omega}{m\omega_0(\omega^2-\omega_0^2)}\frac{\omega_0}{s^2+\omega_0^2}-\frac{F_0}{m(\omega^2-\omega_0^2)}\frac{\omega}{s^2+\omega^2}$$
Thus by taking the inverse transform we obtain $X(t)$
$$X(t)=x_0\cos\omega_0t+\frac{1}{\omega_0}\left[v_0+\frac{F_0\omega}{m(\omega^2-\omega_0^2)}\right]\sin\omega_0t-\frac{F_0}{m(\omega^2-\omega_0^2)}\sin\omega t$$
The motion is the superposition of two simple harmonic motions, one with frequency $\omega_0$ (the natural component of the vibration) and the other with frequency $\omega$ (the forced component of the vibration). The natural vibrations are not present if
$$x_0=0,\ v_0=\frac{F_0\omega}{m(\omega_0^2-\omega^2)}$$
When $\omega=\omega_0$, \begin{align*}x(s)&=\frac{x_0s+v_0}{s^2+\omega_0^2}+\frac{F_0}{m}\frac{\omega_0}{(s^2+\omega_0^2)^2}\\&=\frac{s}{s^2+\omega_0^2}x_0+\frac{v_0}{\omega_0}\frac{\omega_0}{s^2+\omega_0^2}+\frac{F_0}{2m}\frac{1}{s}\frac{2\omega_0s}{s^2+\omega^2} \end{align*} In the first example here, we found $\mathcal{L}^{-1}\left\{\frac{2\omega_0s}{(s^2+\omega_0^2)^2}\right\}=t\sin\omega_0t$. Using the formula here \begin{align*} X(t)&=x_0\cos\omega_0t+\frac{v_0}{\omega_0}\sin\omega_0t+\frac{F_0}{2m}\int_0^t\tau\sin\omega_0\tau d\tau\\&=x_0\cos\omega_0t+\frac{1}{\omega_0^2}\left(v_0\omega_0+\frac{F_0}{2m}\right)\sin\omega_0t-\frac{F_0}{2m\omega_0}t\cos\omega_0t \end{align*} In view of the last term, the amplitude of the oscillations increases indefinitely. In this case, the force $F(t)$ is said to be in resonance with the system.
Vibrations can be quite destructive. A striking example is the collapse of Tacoma Narrows Bridge in 1940. In my college mechanics class I was told that this was resulted from resonance. However the catastrophic vibrations were not actually due to simple mechanical resonance but to a complicated interaction between the bridge and the winds passing through it. Such a phenomenon is called flutter, which is a kind of self-oscillation.
The following is a footage of the old Tacoma Narrows Bridge collapsing:
$y=\sin x$ is not a one-to-one function so there cannot be an inverse function of $y=\sin x$. However we we restrict its domain to the closed interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ it becomes a one-to-one function as shown in Figure 1.
Figure 1. The graph of y=sin(x) on [-pi/2,pi/2]
This means that we can consider $y=\sin^{-1}x$, the inverse function of $y=\sin x$. That is to say, $y=\sin^{-1}x$ is the value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $x=\sin y$. $y=\sin^{-1}x$ is also denoted by $y=\arcsin x$. Similarly,
$y=\cos^{-1}x$ (or $y=\arccos x$) is the value in $[0,\pi]$ for which $x=\cos y$.
$y=\tan^{-1}x$ (or $y=\arctan x$) is the value in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which $x=\tan y$.
$y=\cot^{-1}x$ (or $y=\mathrm{arccot} x$) is the value in $(0,\pi)$ for which $x=\cot y$.
Remark. In general, $y=\sin^{-1}x$ is an inverse relation of $y=\sin x$ which is multiple-valued. In order to consider differentiation, we require it to be single-valued and when $-\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}$ we call it the principal value of $\sin^{-1}x$ and denote it by $\mathrm{Sin}^{-1}x$. Throughout this note we will only consider principal values so we won’t be using the traditional notation like $y=\mathrm{Sin}^{-1}x$.
Recall that the graph of $y=f(x)$ and the graph of its inverse function $y=f^{-1}(x)$ are symmetric about the line $y=x$. Using this symmetry one can obtain the graph of an inverse trigonometric function. For example, Figure 2 shows the graph of $y=\sin x$ and the graph of $y=\sin^{-1} x$.
Figure 2. The graphs of y=sin(x) (in red), y=arcsin(x) (in blue) and y=x (in black).
Figure 3. The graph of y=arcsin(x)
Figure 4. The graph of y=arccos(x)
Figure 5. The graph of y=arctan(x) on [-20,20].
Figure 6. The graph of y=arccot(x) on [-20,20].
In addition, $y=\sec^{-1}x$ has domain $|x|\geq 1$ and range $\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]$.
Figure 7. The graph of y=arcsec(x). The horizontal asymptote is y=pi/2.
$y=\csc^{-1}x$ has domain $|x|\geq 1$ and range $\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$.
The graph of y=arccsc(x).
Inverse Functions and Their Derivatives
Let $y=f^{-1}(x)$ denote the inverse function of $f(x)$. Then \begin{equation}\label{eq:invfn}x=f(y)\end{equation} Differentiate \eqref{eq:invfn} with respect to $x$. \begin{equation}\label{eq:dinvfn}1=f'(y)\frac{dy}{dx}\end{equation} Solving \eqref{eq:dinvfn} for $\frac{dy}{dx}$ we have \begin{equation}\label{eq:dinvfn2}\frac{dy}{dx}=\frac{1}{f'(f^{-1}(x))}\end{equation}
Example. Let $f(x)=\ln x$. Knowing $f'(x)=\frac{1}{x}$ find the derivative of $f^{-1}(x)=e^x$.
Solution. Using \eqref{eq:dinvfn2} $$\frac{d}{dx}e^x=\frac{1}{\frac{1}{e^x}}=e^x$$
Although knowing the formula \eqref{eq:dinvfn2} is convenient, you can always find the derivative of an inverse function by following the same process of deriving \eqref{eq:dinvfn2}. It’s not actually anymore difficult or complicated than using \eqref{eq:dinvfn2}.
The Derivative of Inverse Trigonometric Functions
Let $f(x)=\sin x$. Then $f^{-1}(x)=\sin^{-1}x$. Using \eqref{eq:dinvfn2} \begin{align*}\frac{d}{dx}\sin^{-1}x&=\frac{1}{\cos(\sin^{-1}x)}\\&=\frac{1}{\sqrt{1-\sin^2(\sin^{-1}x)}}\\&=\frac{1}{\sqrt{1-x^2}}\end{align*} where $|x|<1$. The reason the sign in front of $\sqrt{}$ is positive is that $-\frac{\pi}{2}<\sin^{-1}x<\frac{\pi}{2}$ so $\cos(\sin^{-1}x)> 0$.
Alternative derivation: Let $y=\sin^{-1}x$. Then $x=\sin y$. By implicit differentiation $$1=\cos y\frac{dy}{dx}$$ Hence $$\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\cos(\sin^{-1}x)}=\frac{1}{\sqrt{1-x^2}}$$
If $u$ is a functions of $x$, then $$\frac{d}{dx}\sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1$$ Similarly we obtain the rest of derivative formulas. \begin{align*}\frac{d}{dx}\cos^{-1}u&=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1\\\frac{d}{dx}\tan^{-1}x&=\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\sec^{-1}u&=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\\\frac{d}{dx}\csc^{-1}x&=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\end{align*} In turns out we don’t really have to calculate all these formulas. We just need to calculate for example $\frac{d}{dx}\sin^{-1}x$, $\frac{d}{dx}\tan^{-1}x$, $\frac{d}{dx}\sec^{-1}x$ the rest can be obtained by inverse function-inverse cofunction identities \begin{align*}\cos^{-1}x&=\frac{\pi}{2}-\sin^{-1}x\\\cot^{-1}x&=\frac{\pi}{2}-\tan^{-1}x\\\csc^{-1}x&=\frac{\pi}{2}-\sec^{-1}x\end{align*}In case you haven’t seen this identities before they can be easily obtained from cofunction identities. For example sine and cosine are cofunctions of each other as you learned in trigonometry, namely $$\sin\left(\frac{\pi}{2}-x\right)=\cos x,\ \cos\left(\frac{\pi}{2}-x\right)=\sin x$$ Let $$\cos\left(\frac{\pi}{2}-y\right)=\sin y=x$$ Then $$\frac{\pi}{2}-y=\cos^{-1}x$$ i.e. the first identity above $$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$$
Integration Formulas
For any constant $a\ne 0$,
$\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C$, valid for $u^2<a^2$
Example. \begin{align*}\int\frac{dx}{e^{2x}-6}&=\int\frac{du}{u\sqrt{u^2-6}}\ (u=e^x)\\&=\frac{1}{\sqrt{6}}\sec^{-1}\left(\frac{e^x}{\sqrt{6}}\right)+C\end{align*} where $e^x>\sqrt{6}$ or equivalently $x>\ln\sqrt{6}\approx 0.8959$.
Integrating Inverses of Functions
Let $y=f^{-1}(x)$. Then $x=f(y)$ and $dx=f'(y)dy$. So with integration by parts, we have \begin{equation}\begin{aligned}\int f^{-1}(x)dx&=\int yf'(y)dy\\&=yf(y)-\int f(y)dy\\&=xf^{-1}(x)-\int f(y)dy\end{aligned}\label{eq:intinvfn}\end{equation} Using \eqref{eq:dinvfn2} we can also rewrite \eqref{eq:intinvfn} as \begin{equation}\label{eq:intinvfn2}\int f^{-1}(x)dx=xf^{-1}(x)-\int\frac{x}{f'(f^{-1}(x))}dx\end{equation}
Using \eqref{eq:intinvfn}\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x-\sin y+C\\&=x\cos^{-1}x-\sin(\cos^{-1}x)+C\end{align*}Since $x=\cos y$, $$\sin(\cos^{-1}x)=\sin y=\sqrt{1-x^2}$$ (Recall that $0\leq\cos^{-1}x\leq\pi$ so $\sin^{-1}x\geq 0$.) Hence, \begin{equation}\label{eq:intinvcos}\int\cos^{-1}xdx=x\cos^{-1}x-\sqrt{1-x^2}+C\end{equation} Of course one can obtain \eqref{eq:intinvcos} using \eqref{eq:intinvfn2} though the required calculation is a bit longer. The integral \eqref{eq:intinvcos} can be also found without using \eqref{eq:intinvfn} or \eqref{eq:intinvfn2}. Using integration by parts\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}dx\\&=x\cos^{-1}x-\sqrt{1-x^2}+C\end{align*} The rest of the integrals of inverse trigonometric functions are given by\begin{align*}\int\sin^{-1}xdx&=x\sin^{-1}x+\sqrt{1-x^2}+C\\\int\tan^{-1}xdx&=x\tan^{-1}x-\frac{1}{2}\ln(1+x^2)+C\\\int\cot^{-1}xdx&=x\cot^{-1}x+\frac{1}{2}\ln(1+x^2)+C\\\int\sec^{-1}xdx&=x\sec^{-1}x-\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\\\int\csc^{-1}xdx&=x\csc^{-1}x+\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\end{align*}
In this note we study how to solve differential equations with variable coefficients using the Laplace transform. For this we need Derivatives of Transforms. Differentiating
$$f(s)=\int_0^\infty e^{-st}F(t)dt$$
with respect to $s$ we obtain
\begin{align*}
f'(s)&=\int_0^\infty e^{-st}(-tF(t))dt\\
&=\mathcal{L}\{-tF(t)\}
\end{align*}
Continue differentiating to find
\begin{equation}
\label{eq:laplace14}
f^{(n)}(s)=\mathcal{L}\{(-t)^nF(t)\}
\end{equation}
for $n=1,2,\cdots$.
Example. Given that $\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$, find $\mathcal{L}\{t\sin kt\}$.
Solution. Using \eqref{eq:laplace14}
\begin{align*}
\mathcal{L}\{t\sin kt\}&=-\frac{d}{ds}\frac{k}{s^2+k^2}\\
&=\frac{2ks}{(s^2+k^2)^2}.
\end{align*}
The equation \eqref{eq:laplace14} together with transform of derivative formula allows us to transform differential equations with variable coefficients. For example,
\begin{align*}
\mathcal{L}\{t^nX(t)\}&=(-1)^nx^{(n)}(s)\\
\mathcal{L}\{t^2\dot{X}(t)\}&=\frac{d^2}{ds^2}[sx(s)-X(0)]\\
&=\frac{d}{ds}[x(s)+sx'(s)]\\
&=sx^{\prime\prime}(s)+2x'(s)\\
\mathcal{L}\{t\ddot{X}(t)\}&=-\frac{d}{ds}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=-s^2x'(s)-2sx(s)+X(0)
\end{align*}
We are now ready to solve differential equations with variable coefficients.
Example. Find the solution of the problem
$$\ddot{X}(t)+t\dot{X}(t)-X(t)=0,\ X(0)=0,\ \dot{X}(0)=1$$
Solution. The transformed equation is
$$s^2x(s)-1-\frac{d}{ds}[sx(s)]-x(s)=0$$
which can be written as the first-order linear differential equation
$$\frac{d}{ds}x(s)+\left(\frac{2}{s}-s\right)x(s)=-\frac{1}{s}$$
The integrating factor is
$$\mu(s)=e^{\int\left(\frac{2}{s}-s\right)ds}=s^2e^{-\frac{1}{2}s^2}$$
and hence the solution $x(s)$ is
\begin{align*}
x(s)&=\frac{\int\mu(s)\left(-\frac{1}{2}\right)ds}{\mu(s)}\\
&=\frac{-\int se^{-\frac{1}{2}s^2}ds}{s^2e^{-\frac{1}{2}s^2}}\\
&=\frac{1}{s^2}+\frac{C}{s^2}e^{\frac{1}{2}s^2}
\end{align*}
where $C$ is a constant. Since $x(s)\to 0$ as $s\to\infty$, $C$ must be $0$. Therefore $x(s)=\frac{1}{s^2}$ and consequently $X(t)=t$.
Example. Solve Bessel’s equation with index zero
$$t\ddot{X}(t)+\dot{X}(t)+tX(t)=0$$
with the initial condition $X(0)=1$.
Solution. The transformed equation is
$$-\frac{d}{ds}[s^2x(s)-s-\dot{X}(0)]+sx(s)-1-\frac{d}{ds}x(s)=0$$
which simplifies to the first-order separable differential equation
$$(s^2+1)x'(s)+sx(s)=0$$
Performing the integrals
$$\int\frac{dx}{x}=-\int\frac{sds}{s^2+1}$$
we find
\begin{align*}
x(s)&=\frac{C}{\sqrt{s^2+1}}\\
&=\frac{C}{s}\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}\\
&=\frac{C}{2}\sum_{n=0}^\infty\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}\left(\frac{1}{s^2}\right)^n,
\end{align*}
where $C$ is a constant and $s>1$, by the Binomial Theorem.
\begin{align*}
\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{(-1)^n1\cdot 2\cdot 3\cdot 5\cdots(2n-1)}{2^nn!}\\
&=\frac{(-1)^n(2n)!}{(2^nn!)^2}
\end{align*}
So we have
$$x(s)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}$$
Since $x(s)$ is an infinite sum we cannot directly use the linearity of $\mathcal{L}^{-1}$ to obtain $X(t)$. Nonetheless we can show that
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
\begin{align*}
\mathcal{L}\left\{C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}&=C\mathcal{L}\left\{\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}\\
&=C\int_0^\infty e^{-st}\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\int_0^\infty e^{-st}\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\int_0^\infty e^{-st}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\mathcal{L}\{t^{2n}\}\\
&=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}\\
&=x(s)
\end{align*}
By the uniqueness of $\mathcal{L}^{-1}$, we have
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
Since $X(0)=1$, we obtain $C=1$ and hence
$$X(t)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
This series is denoted by $J_0(t)$ i.e.
\begin{equation}
\begin{aligned}
J_0(t&)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\\
&=1-\frac{t^2}{2^2}+\frac{t^4}{2^2\times 4^2}-\frac{t^6}{2^2\times 4^2\times 6^2}+\cdots
\end{aligned}\label{eq:bessel0}
\end{equation}
One can easily show using, for example, the ratio test that the series in \eqref{eq:bessel0} converges for all $t$. We now have the Laplace transform
\begin{equation}
\mathcal{L}\{J_0(t)\}=\frac{1}{\sqrt{s^2+1}}\ (s>1)
\end{equation}
The differential equation
\begin{equation}
\label{eq:besseleqn}
t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-n^2)X(t)=0
\end{equation}
is called the Bessel’s equation of index $n$. The solution $X(t)$ is
$$X(t)=CJ_n(t),\ n=0,1,2,\cdots$$
where
\begin{equation}
\label{eq:besseln}
J_n(t)=\sum_{k=0}^\infty\frac{(-1)^k}{k!(n+k)!}\left(\frac{t}{2}\right)^{n+2k}
\end{equation}
$J_n(t)$, $n=0,1,2,\cdots$ is called the Bessel function of the first kind. There is another solution of the Bessel’s equation in \eqref{eq:besseleqn} which is linearly independent from $J_n(t)$. It is in a pretty horrible form
\begin{equation}
\begin{aligned}
N_n(x)=&\frac{2}{\pi}\left[\ln\left(\frac{x}{2}\right)+\gamma-\frac{1}{2}\sum_{p=1}^n\frac{1}{p}\right]J_n(x)\\
&-\frac{1}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r}\sum_{p=1}^r\left[\frac{1}{p}+\frac{1}{p+n}\right]\\
&-\frac{1}{\pi}\sum_{r=0}^{n-1}\frac{(n-r-1)!}{r!}\left(\frac{x}{2}\right)^{-n+2r}
\end{aligned}\label{eq:neumann}
\end{equation}
where $\gamma$ is the Euler-Mascheroni constant defined by
\begin{align*}
\gamma&=\lim_{n\to\infty}\left(\sum_{m=1}^n\frac{1}{m}-\ln n\right)\\
&\approx 0.57721566\cdots
\end{align*}
$N_n(x)$, $n=0,1,2,\cdots$ is called the Bessel function of the second kind or the Neumann function. Hence the general solution of the Bessel’s equation is given by
$$X(t)=AJ_n(x)+BN_n(x)$$
Those who wish to know more about Bessel functions and Neumann functions may refer to the reference [1] below.
Let us now consider $n=1$ case.
$$t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-1)X(t)=0$$
Using \eqref{eq:laplace14} we obtain
\begin{align*}
\mathcal{L}\{t^2\ddot{X}(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{\ddot{X}(t)\}\\
&=\frac{d^2}{ds^2}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=s^2x^{\prime\prime}(s)+4sx'(s)+2x(s)\\
\mathcal{L}\{t\dot{X}(t)\}&=-\frac{d}{ds}\mathcal{L}\{\dot{X}(t)\}\\
&=-\frac{d}{ds}[sx(s)-X(0)]\\
&=-x(s)-sx'(s)\\
\mathcal{L}\{t^2X(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{X(t)\}\\
&=x^{\prime\prime}(s)
\end{align*}
Hence the transformed equation is
\begin{equation}
\label{eq:laplace15}
(s^2+1)x^{\prime\prime}(s)+3sx'(s)=0
\end{equation}
Let $y(s)=x'(s)$. Then \eqref{eq:laplace15} becomes the separable first-order differential equation
$$(s^2+1)y'(s)+3sy(s)=0$$
which can then be written as
$$\frac{dy}{y}=-\frac{3s}{s^2+1}ds$$
Integrating this we find
$$y(s)=\frac{dx}{ds}=\frac{C_1}{(s^2+1)^{\frac{3}{2}}}$$
Using the trigonometric substitution $s=\tan\theta$, we find
$$x(s)=\int\frac{C_1}{(s^2+1)^{\frac{3}{2}}}ds=\frac{C_1s}{\sqrt{s^2+1}}+C_2$$
Since $\lim_{s\to\infty}x(s)=0$, $C_2=-C_1$. Setting $C_1=C$, we have
\begin{equation}
\begin{aligned}
x(s)&=C\left[\frac{s}{\sqrt{s^2+1}}-1\right]\\
&=C[s\mathcal{L}\{J_0(t)\}-J_0(0)]\\
&=C\mathcal{L}\{J_0′(t)\}
\end{aligned}\label{eq:laplace16}
\end{equation}
Hence,
$$X(t)=CJ_0′(t)$$
From \eqref{eq:besseln} we find $J_1(t)=-J_0′(t)$ so
\begin{equation}
\label{eq:laplace17}
X(t)=-CJ_1(t)
\end{equation}
We can obtain the Laplace transform of $J_1(t)$ using \eqref{eq:laplace16} and \eqref{eq:laplace17}.
\begin{align*}
\mathcal{L}\{J_1(t)\}&=1-\frac{s}{\sqrt{s^2+1}}\\
&=\frac{1}{\sqrt{s^2+1}(\sqrt{s^2+1}+s)}
\end{align*}
References:
[1] Mathematical Methods for Physicists, George Arfken, Third Edition, Academic Press, 1985
