# Integration

18 October 1946

## Methods

1. Substitution
2. By parts
3. Differentiation of a parameter
4. Series expansion
5. Contour integration
6. Numerical methods
7. Special tricks

Methods 3 to 7 are applicable particularly to definite integrals.

## Complex Variable in Substitution

Find $$\int_0^\infty e^{-ax} \cos bx ~\mathrm dx.$$

Since $$\cos bx = (e^{ibx} + e^{-ibx})/2$$, we have

\begin{aligned} \int_0^\infty e^{-ax} \cos bx &= \frac12 \int_0^\infty e^{-(a - ib)x} ~\mathrm dx + \frac12 \int_0^\infty e^{-(a + ib)x} ~\mathrm dx \newline &= \frac12 \left( \frac1{a - ib} + \frac1{a + ib} \right) \newline &= \frac{a}{a^2 + b^2} \end{aligned}

## Differentiation of a Parameter

Find $$\displaystyle \int_0^\infty xe^{-ax} \cos bx ~\mathrm dx.$$

Define $$\displaystyle S(a) = \int_0^\infty e^{-ax} \cos bx ~\mathrm dx = \frac{a}{a^2 + b^2}.$$

Differentiate with respect to $$a$$:

$S’(a) = -\int_0^\infty xe^{-ax} \cos bx = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

The general rule for differentiation with respect to a parameter is \begin{aligned}\frac{d}{d\alpha} \left[ \int_{x_1(\alpha)}^{x_2(\alpha)} f(x, \alpha) ~\mathrm dx\right] &= \int_{x_1}^{x_2} \frac{\partial}{\partial\alpha} f(x, \alpha) ~\mathrm dx \newline &+ \left[\frac{d}{d\alpha} x_2(\alpha)\right]\cdot f(x_2, \alpha) - \left[\frac{d}{d\alpha} x_1(\alpha)\right]\cdot f(x_1, \alpha)\end{aligned}

\begin{aligned} \Delta \int_{x_1}^{x_2} f(x, \alpha) ~\mathrm dx &= \int_{x_1}^{x_2} \Delta f(x, \alpha) ~\mathrm dx \newline &+ \Delta x_2(\alpha) f(x_2, \alpha) - \Delta x_1(\alpha) f(x, \alpha) \end{aligned}

## Multiplication by a Factor

Find $$\displaystyle \int_0^\infty \frac{\sin x}x ~\mathrm dx.$$

\begin{aligned} S(\alpha) &= \int_0^\infty e^{-ax} \frac{\sin x}x ~\mathrm dx \newline S’(\alpha) &= -\int_0^\infty e^{-ax} \sin x ~\mathrm dx = \frac{-1}{1 + \alpha^2} \newline S(\alpha) &= -\arctan \alpha + C \newline \lim_{\alpha \to \infty} S(\alpha) &= 0 = -\frac\pi{2} + C \newline S(\alpha) &= \frac\pi{2} - \arctan \alpha \newline S(0) &= \frac\pi{2} \end{aligned}

## Differentiation under an Integral Sign

\begin{aligned} \int_0^\infty e^{-ax} (1 + x^2)^{-1} ~\mathrm dx &= S(a) \newline \int_0^\infty x^2 e^{-ax} (1 + x^2)^{-1} ~\mathrm dx &= S’’(a) \end{aligned}

\begin{aligned} S(a) + S’’(a) &= \int_0^\infty \left(\frac1{1 + x^2} + \frac{x^2}{1 + x^2}\right) e^{-ax} ~\mathrm dx \newline \frac{d^2S}{da^2} + S &= \frac1a \end{aligned}

This differential equation may be solved for $$S(a)$$.

### Partial Differentiation

\begin{aligned} S(\alpha, \beta) &= \int_0^\infty e^{-\alpha x} \sin \beta x ~\mathrm dx \newline \frac{\partial^2 S}{\partial \beta^2} &= -\int_0^\infty x^2 e^{-\alpha x} \sin \beta x ~\mathrm dx \newline \frac{\partial^2 S}{\partial \alpha^2} &= \int_0^\infty x^2 e^{-\alpha x} \sin \beta x ~\mathrm dx \newline \frac{\partial^2 S}{\partial \beta^2} &= - \frac{\partial^2 S}{\partial \alpha^2} \end{aligned}

The form of $$S$$ may be partly determined by making the substitution $$y = \beta x$$.

\begin{aligned} S(\alpha, \beta) &= \frac1\beta \int_0^\infty e^{-\frac\alpha\beta y} \sin y ~\mathrm dy = \frac1\beta F\left(\frac\alpha\beta\right)\newline \frac\partial{\partial\beta} \left[ \frac1\beta F\left(\alpha\beta\right) \right] &= \frac1\beta F’\left(-\frac\alpha{\beta^2}\right) - \frac{F\left(\frac\alpha\beta\right)}{\beta^2} \newline \frac{\partial^2}{\partial\beta^2}\left[\frac1\beta F\left(\alpha\beta\right)\right] &= -\frac{\alpha}{\beta^3} F’’\left(-\frac{\alpha}{\beta^2}\right) + F’\left(\frac\alpha\beta\right)\frac{3\alpha}{\beta^4} \newline &+ F\left(\frac\alpha\beta\right) \frac2{\beta^3} - \frac1{\beta^2}F’\left(-\frac{\alpha}{\beta^2}\right) \newline \newline \frac{\partial^2S}{\partial\beta^2} &= \frac2{\beta^3}F + \frac{4\alpha}{\beta^4}F’ + \frac{\alpha^2}{\beta^5}F’’ \newline \frac{\partial^2S}{\partial\alpha^2} &= \frac1{\beta^3}F’' \end{aligned}

Multiply through by $$\beta^3$$ and let $$z = \frac{\alpha}{\beta}$$. The original has $$\frac{d}{dx}$$ for $$\frac{d}{dz}$$ in what follows.

\begin{aligned} 2F + 4zF’ + z^2F’’ &= -F’’ \newline \frac{d}{dz}(z^2F’) + 2\frac{d}{dz}(zF) &= -F’’ \newline z^2F’ + 2zF &= -F’ + C_1 \newline \frac{d}{dz}(z^2F) &= -F’ + C_1 \newline z^2F &= -F + C_1z + C_0 \newline \end{aligned}

\begin{aligned} F &= \frac{C_0 + C_1z}{1 + z^2}\newline S(\alpha, \beta) &= \frac1\beta F = \frac{C_0\beta + C_1\alpha}{\alpha^2 + \beta^2} \end{aligned}

To evaluate $$c_0$$ and $$c_1$$, we observe:

$S(\alpha, 0) = 0 = \frac{C_1\alpha}{\alpha_2} \quad\therefore C_1 = 0$

For small $$\beta$$, $$\sin \beta x \sim \beta x$$. This is a good approximation since $$e^{-\alpha x}$$ kills the integrand for large $$x$$.

\begin{aligned} \int_0^\infty e^{-\alpha x} \beta x ~\mathrm dx &= \frac\beta{\alpha^2} \newline \frac{C_0\beta}{\alpha^2 + \beta^2} &\cong \frac{C_0\beta}{\alpha^2} \quad\therefore C_0 = 1 \end{aligned}

$S(\alpha, \beta) = \frac\beta{\alpha^2 + \beta^2}$

## The Dirac Delta Function

$$\delta(x) = 0$$, $$x \not= 0$$, and is in some way infinite for $$x = 0$$. The reference for this section is The Principles of Quantum Mechanics, P. M. Dirac.

$\begin{gathered} \int_{-\infty}^\infty \delta(x) ~\mathrm dx \vcentcolon= \int_{-\epsilon}^{\epsilon} \delta(x) ~\mathrm dx \vcentcolon= 1 \newline \int_{-\infty}^\infty f(x) \delta(x) ~\mathrm dx = f(0) \end{gathered}$

Proof:

Write $\displaystyle \int_{-\infty}^\infty f(x) \delta(x) ~\mathrm dx = \int_{-\infty}^{-\epsilon} f(x) \delta(x) ~\mathrm dx + \int_{-\epsilon}^\epsilon f(x) \delta(x) ~\mathrm dx + \int_\epsilon^\infty f(x) \delta(x) ~\mathrm dx.$

If $$f(x)$$ varies relatively slowly in the range $$[-\varepsilon, \varepsilon]$$, we can replace it by an average of $$f(x)$$ over that interval.

\begin{aligned} \int_{-\infty}^\infty f(x) \delta(x) ~\mathrm dx &= \int_{-\epsilon}^\epsilon f(x) \delta(x) ~\mathrm dx \newline &= \int_{-\epsilon}^\epsilon \overline{f(x)} \delta(x) ~\mathrm dx \newline &= \overline{f(x)} \cdot \mathbf{1}{x = 0} \cdot \int{-\epsilon}^\epsilon \delta(x) ~\mathrm dx \newline \left(\lim_{\epsilon \to 0}\right) \quad &= f(0). \end{aligned}

The delta function is useful since it may be operated on as though it were a real function. The only justification we give for this is that it gives the correct answer.

\begin{aligned} f’(t) &= -\int_{-\infty}^\infty f(x) \delta’(x - t) ~\mathrm dx \newline &= \left. f(x) \delta(x - t) \right\vert_{x=-\infty}^{x=\infty} - \int_{-\infty}^\infty f’(x) \delta(x - t) ~\mathrm dx \newline &= f’(t). \end{aligned}

The following useful relations may be easily proved:

1. $$\delta(x) = \delta(-x)$$
2. $$\delta’(x) = -\delta’(-x)$$
3. $$x \delta(x) \equiv 0$$
4. $$x\delta’(x) = -\delta(x)$$
5. $$\delta(ax) = \frac1{\vert a\vert} \delta(x)$$
6. $$f(x)\delta(x - a) = f(a)\delta(x - a)$$

The following figures are approximate representations of $$\delta(x)$$ and $$\delta’(x)$$.

Define the step function There are differing definitions of the value at $$x = 0$$. The definition of $$\delta(x)$$ as the limit $$\sigma \to 0$$ of the normal distribution suggests the value should be $$1/2$$, but some authors define it as $$1$$. $$\displaystyle H(x) = \int_{-\infty}^x \delta(x) ~\mathrm dx = \begin{cases} 0, &x < 0,\newline 1, &x >0.\end{cases}$$

Consider the piecewise smooth curve above. Its derivative may be expressed in terms of the delta function as $$f’(x)_{x \not= a} + c\delta(x - a).$$

Consider the integral $$\int_{-\infty}^\infty \cos \beta x ~\mathrm dx$$. We shall show that it is equal to $$2\pi \delta(\beta)$$ by showing $\int_0^\infty \cos \beta x ~\mathrm dx = \pi \delta(\beta).$

Note You may argue here that we haven't shown that this limit is actually equal to some multiple of $$\delta(\beta)$$. You'd be correct to do so. The fact Feynman is implicitly using here is that the Dirac delta is the limit of a Cauchy distribution as the scale parameter goes to $$0$$. In fact, Cauchy defined the delta this way 100 years before Dirac! $$\displaystyle\lim_{\alpha \to 0} e^{-\alpha x} \cos \beta x ~\mathrm dx = \lim_{\alpha \to 0} \frac\alpha{\alpha^2 + \beta^2}.$$

For small $$\alpha$$ this integral looks approximately as above. The area under the curve is $$A = \int_{-\infty}^\infty \frac\alpha{\alpha^2+\beta^2} ~\mathrm d\beta.$$

Making the change of variables $$\beta = \alpha z$$, we obtain

\begin{aligned} A &= \int_{-\infty}^\infty \frac{\alpha^2}{\alpha^2 + \alpha^2 z^2} ~\mathrm dz \newline &= \int_{-\infty}^\infty \frac{dz}{1 + z^2} \newline &= \arctan z \vert_{-\infty}^\infty \newline &= \pi. \end{aligned}

Note that the area is independent of the value of $$\alpha.$$ And thus $$A$$ is equal to the multiple of $$\delta(\beta)$$ whose integral over the real line is $$\pi,$$ which is $$\pi \delta(\beta).$$

### Example: Use of the delta function in evaluating a definite integral

$\int_0^{\pi/2} \cos(m \tan\theta) ~\mathrm d\theta$

Substitute $$\tan \theta = x$$.

\begin{aligned} \int_0^\infty \cos mx ~\frac{dx}{1 + x^2} &= S(m) \newline -\int_0^\infty x^2 \cos mx ~\frac{dx}{1 + x^2} &= S’’(m) \newline \end{aligned}

$S(m) - S’’(m) = \int_0^\infty \cos mx ~\mathrm dx = \pi \delta(m)$

### Aside: Variation of parameters

This is a perfectly general method for solving inhomogeneous differential equations. It shows up in any undergraduate course, but this example begs for more details here, so let's go through it now.

In fact, let's solve the more general equation $$S’’(m) - S(m) = f(m)$$. First, we find the homogeneous solutions $$h_1(m) = e^m$$, $$h_2(m) = e^{-m}.$$ These are so named because they solve the homogeneous equation $$S’’(m) - S(m) = 0.$$

Now let's assume the solution is given by This is why the method is called variation of parameters; the "parameters" are the "coefficients" on $$h_1$$ and $$h_2$$, which would be constants were this equation homogeneous; but we've just made them variables. $S(m) = s_1(m)h_1(m) + s_2(m)h_2(m)$ and further assume Yes, there is a valid reason we can make all these assumptions; it has to do with something called differential algebra, which explains how the derivative acts as a linear operator on a function space. $0 = s_1’(m)h_1(m) + s_2’(m)h_2(m).$

If we differentiate the foregoing equation, we get \begin{aligned} S’(m) &= s_1’(m)h_1(m) + s_1(m)h_1’(m) \newline &\quad + s_2’(m)h_2(m) + s_2(m)h_2’(m) \newline &=s_1(m)h_1’(m) + s_2(m)h_2’(m). \end{aligned}

If we do it again, we get

\begin{aligned} S’’(m) &= s_1’(m) h_1’(m) + s_1(m) h_1’’(m) \newline &\quad+ s_2’(m) h_2’(m) + s_2(m) h_2’’(m). \end{aligned}

Now we have \begin{aligned} f(m) = S’’(m) - S(m) &= s_1’(m)h_1’(m) + s_1(m)h_1’’(m) \newline &\quad + s_2’(m)h_2’(m) + s_2(m)h_2’’(m) \newline &\quad - s_1(m)h_1(m) - s_2(m)h_2(m) \newline &= s_1’(m)h_1’(m) + s_2’(m)h_2’(m). \end{aligned}

Thus we obtain the system of equations This trick of "constructing $$f(m)$$" is the key step here; it goes through regardless of what the original equation was. Everything after this is problem-specific.

$\begin{bmatrix} h_1(m) & h_2(m) \newline h_1’(m) & h_2’(m) \end{bmatrix} \begin{bmatrix} s_1’(m) \newline s_2’(m) \end{bmatrix} = \begin{bmatrix} 0 \newline f(m)\end{bmatrix}$

In our case, this becomes

\begin{aligned} e^m s_1’(m) + e^{-m} s_2’(m) &= 0, \newline e^ms_1’(m) - e^{-m} s_2’(m) &= -\pi\delta(m),\end{aligned}

which gives $$2e^m s_1’(m) = -\pi \delta(m)$$, $2e^{-m}s_2’(m) = \pi\delta(m)$, so we have

\begin{aligned} s_1(m) &= -\frac\pi2 \int e^{-m} \delta(m) ~\mathrm dm \newline &= -\frac\pi2 e^{-0}H(m) + A, \newline s_2(m) &= \frac\pi2 \int e^m \delta(m) ~\mathrm dm \newline &= \frac\pi2 e^0H(m) + B. \end{aligned}

Thus the solution is given by

\begin{aligned} S(m) &= \left(-\frac\pi2 H(m) + A\right) e^m + \left(\frac\pi2 H(m) + B\right)e^{-m} \newline &= \frac\pi2 H(m) (-e^m + e^{-m}) + Ae^m + Be^{-m}, \end{aligned}

which agrees with what is given below.

### Example continued

The general solution of such a differential equation is Obtained by variation of parameters.

$\displaystyle S(m) = \frac{e^m}2 \int_{-\infty}^m e^{-t} f(t) ~\mathrm dt - \frac{e^{-m}}2\int_{-\infty}^m e^t f(t) ~\mathrm dt + Ae^m + Be^{-m}$

where $$f(t) = \pi \delta(t)$$.

Therefore the solution has the form

$S(m) = \begin{cases}Ae^m + Be^{-m}, &m < 0, \newline Ae^m + Be^{-m} - \frac\pi2 e^m + \frac\pi2 e^{-m}, &m > 0,\end{cases}$

where the constants $$A$$, $$B$$ still have to be determined. Note $$S(0) = \pi/2 = A + B$$. Also note that $$S(m)$$ is an even function, Note that $$T(\theta) = \int \cos(m \tan \theta) ~\mathrm dm$$ would be odd. so we have After assuming without loss of generality that $$m > 0$$.

\begin{aligned} S(-m) &= S(m) \newline Ae^{-m} + Be^m &= Ae^m + Be^{-m} - \frac\pi2 e^m + \frac\pi2 e^{-m} \newline (B - A)(e^{-m} - e^m) &= -\frac\pi2 (e^{-m} - e^m) \newline B - A &= -\frac\pi2, \end{aligned}

from which it follows that $$A = \pi/2$$ and $$B = 0$$, and we finally obtain

\begin{aligned} S(m) &= \begin{cases} \frac\pi2 e^m, &m < 0, \newline \frac\pi2 e^{-m}, &m \geq 0.\end{cases} \end{aligned}

## More About the Delta Function

Let us consider the integral

\begin{aligned} S(\beta) &= \int_{-\infty}^\infty \frac{\sin \beta x}x dx \newline S’(\beta) &= \int_{-\infty}^\infty \frac{x \cos \beta x}x dx = 2\pi \delta(\beta)\newline S(\beta) &= \begin{cases} 2\pi + c, &\beta > 0, \newline c, &\beta < 0.\end{cases} \end{aligned}

[Figure goes here]

Consider now the integral Evaluating this is a nice, quick exercise, given previous results. Note that a little work and the right choice of domain shows $\frac{\pi}2 - \arctan \frac\alpha\beta = \arctan \frac\beta\alpha.$

$\int_0^\infty e^{-\alpha x} \frac{\sin \beta x}x ~\mathrm dx = \frac{\pi}{2} - \arctan \frac{\alpha}{\beta}$

A graph of this integral looks as follows:

[Figure]

In the $$\lim_{a\to 0},$$ this integral reduces to the one considered previous to it. (There is a factor of 2 to be considered.)

## Investigation of the Existence of Integrals

Consider $$\displaystyle \int_0^\infty \frac{e^{-ax}}x ~\mathrm dx.$$

Near $$0$$, this integral is approximately $$\int_0^\infty \frac1x ~\mathrm dx = -\ln 0 + \dots.$$ The integral is infinite.

Consider $$\displaystyle \int_0^\infty \frac{1 - e^{-ax}}x ~\mathrm dx.$$

As $$x \to 0$$, the integral is approximately $$\int_0^\infty a ~\mathrm dx = 0 + \dots$$.

As $$x \to \infty$$, the integral is approximately $$\int_0^\infty \frac1x ~\mathrm dx = \ln\infty + \dots.$$ Again, the integral is infinite.

The following table of functions may prove useful in determining whether or not a given integral exists.

Function Finite as $$x \to \infty$$ Finite as $$x \to 0$$
$$dx/x^2$$ Yes No
$$dx/x$$ No No
$$dx/x^{1 + \epsilon}$$ Yes No
$$dx/x^{1 - \epsilon}$$ No Yes
$$dx$$ No Yes

## Special Method of Evaluating a Definite Integral

Let $$\displaystyle A = \int_{-\infty}^\infty e^{-x^2} ~\mathrm dx$$. This is the probability integral, so named because it's the (scaled) pdf of the normal distribution.

This integral exists, Because $$e^{-x^2} < e^{-\vert x\vert}$$ if $$\vert x \vert > 1$$ and $$\int_{-\infty}^\infty e^{-\vert x \vert} ~\mathrm dx = 2$$. so its square exists: The last line works because the second integral is in a different variable than, and thus constant with respect to, the first.

\begin{aligned} A^2 &= \int_{-\infty}^\infty e^{-x^2} ~\mathrm dx \cdot \int_{-\infty}^\infty e^{-y^2} ~\mathrm dy \newline &= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} ~\mathrm dx ~\mathrm dy. \end{aligned}

Now we convert the double integral to polar coordinates. Make the change of variables $$x = r \cos \theta$$, $$y = r \sin \theta$$: \begin{aligned} &= \int_0^\infty \int_0^{2\pi} e^{-r^2} \cdot r ~\mathrm d\theta ~\mathrm dr \newline &= \int_0^\infty re^{-r^2} \theta\vert_{\theta=0}^{\theta=2\pi} ~\mathrm dr \newline &= \int_0^\infty 2\pi re^{-r^2} ~\mathrm dr \newline &= 2\pi \lim_{c \to \infty} \left. -\frac12 e^{-r^2} \right\vert_0^c \newline &= -\pi \lim_{c \to \infty} e^{-c^2} - e^0 \newline &= \pi, \end{aligned} whereupon we obtain $$A = \sqrt\pi.$$

## Series Evaluation of Definite Integrals NB: This section was originally titled "Series Solution of Differential Equations," but that doesn't fit.

Let's compute $$\displaystyle \int_0^\infty \frac{dx}{e^{nx} + e^{-nx}}.$$$$= \int_0^\infty \frac12 \operatorname{sech} nx ~\mathrm dx.$$

We have

\begin{aligned} \int_0^\infty \frac{dx}{e^{nx} + e^{-nx}} &= \int_0^\infty \frac{e^{nx}}{e^{2nx} + 1} ~\mathrm dx\newline \end{aligned}

This integral can now be evaluated immediately with the substitution $u = e^{nx}$, but let's do it another way by changing it to a series:

\begin{aligned} &= \int_0^\infty e^{nx} \cdot \frac{dx}{1 - (-e^{2nx})}. \end{aligned}

Since $$e^{-2nx} < 1$$ if $$x > 0$$, we have \begin{aligned} &= \int_0^\infty e^{nx} \left(\sum_{k = 0}^\infty (-1)^k e^{-2knx}\right) ~\mathrm dx \newline &= \int_0^\infty e^{nx} \left(\sum_{k = 0}^\infty (-1)^k e^{-2knx}\right) ~\mathrm dx, \end{aligned} and since the series converges uniformly, we can switch the integral and summation, giving \begin{aligned} &= \sum_{k = 0}^\infty \int_0^\infty (-1)^k e^{(-2k + 1)nx} ~\mathrm dx \newline &= \sum_{k = 0}^\infty (-1)^k \left. \frac{1}{(-2k + 1)n} \cdot e^{(-2k + 1)nx} \right\vert_0^\infty \newline &= \sum_{k = 0}^\infty \frac{(-1)^k}{(-2k + 1)n} \newline &= \frac1n \sum_{k = 0}^\infty \frac{(-1)^k}{(-2k + 1)}. \end{aligned} Then, if we observe that the series is equal to the Taylor series for $$\arctan x$$ evaluated at $$x = 1$$, we can write $\boxed{\int_0^\infty \frac{dx}{e^{nx} + e^{-nx}} = \frac1n \arctan 1 = \frac1n \cdot \frac\pi4.}$

A combination of the methods of integration discussed thus far is sufficient to solve any integral which exists. There are still other methods which may be used, and are in many cases easier. We shall discuss some of them later.

## Problems

1. Given $$\displaystyle\int_0^\infty e^{-a^2x^2} ~\mathrm dx = \frac{\sqrt\pi}{2a},$$ prove that $\int_0^\infty e^{-a^2x^2} \cos bx ~\mathrm dx = \frac{\sqrt\pi}{2a} e^{-\frac{b^2}{4a^2}}.$

2. Given $$\displaystyle\int_0^\infty e^{-x^2} ~\mathrm dx = \sqrt\pi/2$$, find $\displaystyle\int_0^\infty x^4e^{-x^2} ~\mathrm dx.$

3. Find $\int_0^\infty \frac{e^{-ay} - e^{-by}}y ~\mathrm dy.$

4. Find $\int_{-\infty}^\infty e^{-\beta x^2 - \frac\alpha{x^2}} ~\mathrm dx.$

5. Find $\int_0^\infty \frac{\sin^2 x}{x^2} ~\mathrm dx.$

6. Prove $$\delta(ax) = \frac1a \delta(x)$$ for $$a > 0$$, in the sense that they have equal integrals over any interval.

7. Find another form for $$x\delta’’(x)$$.

8. FindHint: Use the previous derivation to obtain $\int_0^\infty \frac{\cos mx}{k^2 + x^2} ~\mathrm dx.$ Then differentiate with respect to $$k$$. $\displaystyle \int_0^\infty \frac{\cos mx}{(1 + x^2)^2} ~\mathrm dx.$

9. Show that These are called the Fresnel integrals, and they are demonstrated beautifully in the Cornu spiral. $\displaystyle \int_0^\infty \cos(x^2) ~\mathrm dx = \int_0^\infty \sin(x^2) ~\mathrm dx = \frac12 \sqrt{\frac\pi2}.$

10. Find $\int_0^\infty \frac{x}{e^{mx} - e^{-mx}} ~\mathrm dx.$

11. Let $$\displaystyle\int_0^\infty \frac{\sin x}{\sinh ax} ~\mathrm dx = S(a).$$

1. Find $$S(1)$$ to 3 significant figures.
2. Find approximate expressions for $$S(a)$$ for large and small $$a$$.