# <la-tex> Custom Element

## Note

• The source code of this page contains <la-tex> custom elements with LaTeX expressions.
• This demo relies on MathML support to render various mathematical formulas.

## Basic Usage

The <la-tex> tag is a container for LaTeX expressions. The source is automatically converted into MathML using TeXZilla and the mathematical formula is displayed by your rendering engine. In particular, you need a standard-compliant HTML5 rendering engine as well as appropriate math fonts in order to get a good rendering.

You can use the <la-tex> tag to write inline mathematical expressions such as \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. The <la-tex> tag accepts the display and dir attributes of the math element. For instance you can write <la-tex display="block"> to insert display mathematical expressions such as

\Gamma(t) = \lim_{n \to \infty} \frac{n! \; n^t}{t \; (t+1)\cdots(t+n)}= \frac{1}{t} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^t}{1+\frac{t}{n}} = \frac{e^{-\gamma t}}{t} \prod_{n=1}^\infty \left(1 + \frac{t}{n}\right)^{-1} e^{\frac{t}{n}}

or <la-tex display="rtl"> to write right-to-left mathematical expressions such as س = \frac{-ب\pm\sqrt{ب^٢-٤اج}}{٢ا}. Note that the previous expression uses Arabic characters. In general you can use any Unicode characters in the LaTeX source: for example <la-tex>∑_{n=1}^{+∞} \frac{1}{n^2} = \frac{π^2}{6}</la-tex> will render as ∑_{n=1}^{+∞} \frac{1}{n^2} = \frac{π^2}{6}. If a parsing error occurs, then the original LaTeX source is displayed: \gamma + \frac{x}.

Just like the [itex] tag, the <la-tex> tag can be used in other languages from the HTML5 family as shown below. Note that clicking on the HTML button below will retrieve the original LaTeX source, via the .textContent property of the <la-tex> element.

Finally, the <la-tex> tag will react to dynamic changes, as shown in the following example where we use Javascript to modify the display and dir attributes of the <la-tex> tag or to set its .textContent property. ; ; \sqrt 2

## Sample Document using <la-tex>

      Copyright (C) 2013 Frédéric Wang

Permission is granted to copy, distribute and/or modify this document
under the terms of the GNU Free Documentation License, Version 1.3
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts.
A copy of the license is included in the section entitled "GNU


### Basic Integration

#### Ideas

If F is a primitive of f then \int_{a}^b f(x) d x = F(b) - F(a), provided the integrand and integral have no singularities on the path of integration. As a consequence, we can use a table of derivatives:

 Function Derivative Linearity a f + b g a f' + b g' Leibniz rule f g f' g + g f' Reciprocal rule \frac{1}{f} -\frac{f'}{f^2} Chain Rule f \circ g (f' \circ g) g' Inverse function rule f^{-1} \frac{1}{f' \circ f^{-1}} Elementary power rule x^n n x^{n-1} Generalized power rule f^g = e^{g \ln(f)} f^g \left( f' \frac{g}{f} + g' \ln(f) \right) Exponential \exp x \exp x Logarithm \ln x \frac{1}{x} Sine \sin x \cos x Cosine \cos x -\sin x Tangent \tan x \frac{1}{\cos^2 x} = 1+\tan^2 x

#### Examples

Using linearity and elementary power rule:

\int_0^{1} u^4 - 2u^3 + 5u^2 + 4 du = \left[ \frac{u^5}{5} - 2 \frac{u^4}{4} + 5 \frac{u^3}{3} + 4u \right]_{u=0}^1 = \frac{161}{30}

Using linearity and sine/cosine:

\int_0^{\pi} 2 \cos(\theta) - 3 \sin(\theta) d\theta = \left[ 2 \sin(\theta) + 3 \cos(\theta) \right]_{\theta=0}^\pi = -6

Using Leibniz rule, Chain rule and Exponential/Power:

\int_{0}^{2} 2x e^{-3x} - 3 x^2 e^{-3x} d x = \left[ x^2 e^{-3x} \right]_0^2 = \frac{4}{e^{12}}

Using the inverse function rule and tangent:

\int_{0}^{1} \frac{1}{1 + x^2} d x = \int_{0}^{1} \frac{1}{\tan'(\arctan(x))} d x = \left[ \arctan(x) \right]_0^1 = \frac{\pi}{4}

Using linearity, reciprocal rule and logarithm:

\int_{2}^{3} \frac{1}{v (\ln(v))^2} d v = - \int_{2}^{3} - \frac{\frac{1}{v}}{(\ln(v))^2} d v = - \left[ \frac{1}{\ln(v)} \right]_2^3 = \frac{1}{\ln 2} - \frac{1}{\ln 3}

Using the power rules and chain rule:

\int_{0}^1 \frac{3t^2}{(t^3+1)^5} d t = -\frac{1}{4} \int_0^1 (t^3+1)^{-4} \left( 3t^2 \frac{-4}{t^3+1} + 0 \right) d t = -\frac{1}{4} \left[ (t^3+1)^{-4} \right]_0^1 = \frac{15}{64}

### Integration by Substitution

#### Idea

Do a substitution x = g(y) to simplify an integral. We have \frac{dx}{dy} = g'(y) and \int f(x) dx = \int f(g(y)) g'(y) dy: this is the chain rule!

#### Example 1

\int_{0}^1 x (x^2 - 1) dx

Let y = x^2 - 1, dy = 2 x dx then we get

\frac{1}{2} \int_{y=-1}^0 y dy = \left[\frac{y^2}{4}\right]_{y=-1}^0 = -\frac{1}{4}

We find the same result as

\int_{0}^1 x^3 - x dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{x=0}^1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}

#### Example 2

Consider

\int_2^3 6x^2 \cos(x^3 - 2) d x

Let y = x^3 - 2, dy = 3x^2 dx so

\int_{2}^{3} 6 x^2 \cos(x^3 - 2) d x = 2 \int_{y=6}^{25} \cos(y) d y = 2 [\sin(y)]_{y=6}^{25} = 2 \sin(25) - 2 \sin(6)

#### Example 3

Consider

\int_{v=1}^2 \frac{dv}{v^3 e^{\frac{1}{v^2}}}

Let u = -\frac{1}{v^2}, du = \frac{2}{v^3} dv:

\int_{v=1}^2 \frac{dv}{v^3 e^{\frac{1}{v^2}}} = \frac{1}{2} \int_{u=-1}^{-\frac{1}{4}} e^{u} du = \frac{1}{2} \left(\frac{1}{\sqrt[4]{e}} - \frac{1}{e}\right)

#### Example 4

Consider

\int_{-\sqrt{2}}^{-1} \frac{5t}{\sqrt{4 - t^2}} d t

Let t = 2 \sin \theta and so dt = 2 \cos \theta d \theta. We have

\int_{\frac{3\pi}{4}}^{\frac{7\pi}{6}} \frac{10 \sin \theta}{\sqrt{4 - 4\sin^2 {\theta}}} 2 \cos \theta d \theta = 10 \int_\frac{3\pi}{4}^{\frac{7\pi}{6}} \sin \theta \frac{\cos \theta}{\sqrt{\cos^2 \theta}} d \theta

We have -1 < \cos \theta < 0 on \left[\frac{3\pi}{4},\frac{7\pi}{6}\right] and so

\int_{-\sqrt{2}}^{-1} \frac{5t}{\sqrt{4 - t^2}} d t = -10 \int_\frac{3\pi}{4}^{\frac{7\pi}{6}} \sin \theta d \theta = 10 \left[\cos \theta\right]_\frac{3\pi}{4}^{\frac{7\pi}{6}} = 5 (\sqrt{2} - \sqrt{3})

#### Example 5

Consider

\int_{\frac{3}{2}}^{+\infty} \frac{dx}{2x^2 - 6x + 7}

We first write

2x^2 - 6x + 7 = 2\left(x^2 - 3x + \frac{7}{2}\right) = 2\left( \left(x - \frac{3}{2}\right)^2 + \frac{5}{4} \right) = \frac{5}{2} \left( \left(\frac{x - \frac{3}{2}}{\sqrt{\frac{5}{4}}}\right)^2 + 1 \right)

Now doing the substitution y = \frac{x - \frac{3}{2}}{\sqrt{\frac{5}{4}}}, dy = \frac{dx}{\sqrt{\frac{5}{4}}} we get

\int_0^{+\infty} \frac{\sqrt{\frac{5}{4}} dy}{\frac{5}{2}(y^2+1)} = \frac{1}{\sqrt{5}} \int_0^{+\infty} \frac{dy}{y^2+1}

But we have already met this integral in the basic methods:

\int_{\frac{3}{2}}^{+\infty} \frac{dx}{2x^2 - 6x + 7} = \frac{1}{\sqrt{5}} [\arctan y]_{y=0}^{\infty} = \frac{\pi}{2 \sqrt{5}}

#### Example 6

Consider

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan\left(\frac{\theta}{2}\right) \frac{{(1 + \cos \theta)}^2}{\sin^3 \theta} d \theta

We do the substitution t = \tan\left(\frac{\theta}{2}\right). We have d \theta = \frac{2 dt}{1+t^2}, \cos \theta = \frac{1-t^2}{1+t^2} and \sin \theta = \frac{2t}{1+t^2}. Also, we have

\tan\left(\frac{\pi}{12}\right) = \frac{1-\cos\left(\frac{\pi}{6}\right)}{1-\sin\left(\frac{\pi}{6}\right)} = 2 - \sqrt{3}

and so

\int_{2 - \sqrt{3}}^{\frac{1}{\sqrt{3}}} t \frac{\left(\frac{2}{1+t^2}\right)^2}{\left( \frac{2t}{1+t^2}\right)^3} \frac{2 dt}{1+t^2} = \int_{2 - \sqrt{3}}^{\frac{1}{\sqrt{3}}} \frac{1}{t^2} dt = \left[ -\frac{1}{t} \right]_{2 - \sqrt{3}}^{\frac{1}{\sqrt{3}}} = \frac{1}{2-\sqrt{3}}-\sqrt{3}

and finally

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan\left(\frac{\theta}{2}\right) \frac{{(1 + \cos \theta)}^2}{\sin^3 \theta} d \theta = 2

### Integration by parts

#### Idea

Given a two function f, g we have

\int f g' = f g - \int f' g

this is Leibniz rule!

#### Example 1

Consider

\int_{0}^{+\infty} (x^2 + x + 1) e^{-3x} d x

We consider f(x) = x^2 + x + 1 and g(x) = \frac{e^{-3x}}{-3} that is f'(x) = 2x+1 and g'(x) = e^{-3x}. The integral can be written \int f g' dx and hence we get

\left[(x^2 + x + 1) \frac{e^{-3x}}{-3}\right]_{0}^{+\infty} - \int_{0}^{+\infty} (2x+1) \frac{e^{-3x}}{-3} d x = \frac{1}{3} + \frac{1}{3} \int_{0}^{+\infty} (2x+1) e^{-3x} d x

We now consider h(x) = 2x+1, h'(x)=2 the second integral can be written \int h g' dx and hence

\int_{0}^{+\infty} (2x+1) e^{-3x} d x = \left[(2x + 1) \frac{e^{-3x}}{-3}\right]_{0}^{+\infty} - \int_{0}^{+\infty} 2 \frac{e^{-3x}}{-3} d x = \frac{1}{3} + \frac{2}{3} \int_{0}^{+\infty} e^{-3x} d x

We now recognize g' in the last integral and so

\int_{0}^{+\infty} e^{-3x} d x = \left[ \frac{e^{-3x}}{-3}\right]_{0}^{+\infty} = \frac{1}{3}

Finally,

\int_{0}^{+\infty} (x^2 + x + 1) e^{-3x} d x = \frac{1}{3} + \frac{1}{3} \times \left( \frac{1}{3} + \frac{2}{3} \times \frac{1}{3} \right) = \frac{14}{27}

#### Example 2

\int_1^2 x^3 (2\ln x + 7\arctan x) d x

We let f(x) = 2\ln x + 7\arctan x and g(x) = \frac{x^3}{3}. Hence f'(x) = \frac{2}{x} + \frac{7}{1+x^2} and g'(x) = x^3. The integration by parts gives:

\left[\frac{x^3}{3} \left(2\ln x + 7\arctan x\right) \right]_1^2 - \int_1^2 \frac{x^3}{3} \left(\frac{2}{x} + \frac{7}{1+x^2}\right) d x

The integral can be written

\int_1^2 \frac{2}{3} x^2 + \frac{7}{3} \frac{x(x^2+1) - x}{x^2+1} dx = \int_2^2 \frac{2}{3} x^2 + \frac{7}{3} x - \frac{7}{6} \frac{2x}{x^2+1} dx

Finally

\int_1^2 x^3 (2\ln x + 7\arctan x) d x = \left[\frac{x^3}{3} \left(2\ln x + 7\arctan x\right) - \frac{2}{9} x^3 -\frac{7}{6} x^2 + \frac{7}{6} \ln (x^2 + 1) \right]_1^2 = \frac{28}{9} \ln(5) + \frac{89}{18} \ln(2) + \frac{56}{3} \arctan(2) - \frac{7 \pi}{12} - \frac{301}{18}

#### Example 3

Consider

I =\int_{0}^{\frac{\pi}{2}} \sin(3 \theta) e^{2 \theta} d \theta

We do an integration by parts with f(\theta) = \sin(3 \theta) and g'(\theta) = e^{2 \theta}:

I = \left[\sin(3 \theta) \frac{e^{2 \theta}}{2}\right]_{0}^{\frac{\pi}{2}} - \frac{3}{2} \int_{0}^{\frac{\pi}{2}} \cos(3 \theta) e^{2 \theta} d \theta

We repeat a similar integration by parts on the remaining integral:

\int_{0}^{\frac{\pi}{2}} \cos(3 \theta) e^{2 \theta} d \theta = \left[\cos(3 \theta) \frac{e^{2 \theta}}{2}\right]_{0}^{\frac{\pi}{2}} + \frac{3}{2} \int_{0}^{\frac{\pi}{2}} \sin(3 \theta) e^{2 \theta} d \theta

We note that we found the initial integral I. We get:

I = \left[\sin(3 \theta) \frac{e^{2 \theta}}{2} -\frac{3}{4} \cos(3 \theta) e^{2 \theta} \right]_{0}^{\frac{\pi}{2}} - \frac{9}{4} I

Finally,

I =\int_{0}^{\frac{\pi}{2}} \sin(3 \theta) e^{2 \theta} d \theta = \frac{4}{13} \left[\left(\frac{1}{2} \sin(3 \theta) -\frac{3}{4} \cos(3 \theta)\right) e^{2 \theta} \right]_{0}^{\frac{\pi}{2}} = \frac{3 - 2 e^\pi}{13}

### Integration by Partial Fraction Decomposition

#### Idea

Given a rational function \frac{P}{Q}, write its partial fraction decomposition and integrate each term.

#### Example 1

\int_{x=2}^{3} \frac{2x^4-6x^3+41x^2-44x+12}{6x^5-18x^4+18x^3-6x^2} d x

Clearly, we have 6x^5-18x^4+18x^3-6x^2 = 6 x^2 (x^3-3x^2+3x-1) and 1 is a trivial root of the second factor so we can factor it by (x-1) and obtain:

6x^5-18x^4+18x^3-6x^2 = 6 x^2 (x-1) (x^2-2x+1) = 6x^2 (x-1)^3

Hence we search a decomposition of

F(x) = \frac{2x^4-6x^3+41x^2-44x+12}{6 x^2 (x^3-3x^2+3x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}

for some constants A, B, C, D, E \in \mathbb{R}. We have

\lim_{x \rightarrow 1} (x - 1)^3 F(x) = E = \frac{2 - 6 + 41 - 44 + 12}{6} = \frac{5}{6} \lim_{x \rightarrow 0} x^2 F(x) = B = \frac{12}{-6} = -2 \lim_{x \rightarrow 0} (x^2 F(x))' = A = \frac{-44 \times -1 - 12 \times 3}{6} = \frac{4}{3} \lim_{x \rightarrow +\infty} x F(x) = A + C = \frac{2}{6} \rightarrow C = \frac{2}{6} - \frac{4}{3} = -1 F(2) = \frac{A}{2} + \frac{B}{4} + C + D + E = \frac{2\times16 - 6\times8+41\times4-44\times2+12}{6 \times 4} = 3 \rightarrow D = 3 - \frac{2}{3} + \frac{1}{2} + 1 - \frac{5}{6} = 3

Finally,

\int_{x=2}^{3} F(x) d x = \left[ \frac{4}{3} \ln |x| + \frac{2}{x} - \ln |x - 1| - \frac{3}{x-1} - \frac{5}{12} \frac{1}{(x-1)^2} \right]_{x=2}^3 = \frac{4}{3} \ln(3) - \frac{7}{3} \ln(2) + \frac{71}{48}

#### Example 2

\int_{u=1}^{5} \frac{2u^4-5u^3+24u^2-26u+50}{u^5-4u^4+14u^3-20u^2+25u} d u

The denominator can be written u^5-4u^4+14u^3-20u^2+25u = u(u^4-4u^3+14u^2-20u+25). Evaluating the second factor at 0, \pm1, \pm2 we don't find any trivial roots. Hence we try to convert it to a depressed quartic by setting u = v - \frac{-4}{4} = v + 1. We find (u^4-4u^3+14u^2-20u+25) = v^4 + 8v^2 + 16 = (v^2 + 4)^2 = (u^2-2u+5)^2. The discriminant of u^2-2u+5 is 4(1-5) < 0 so we have the decomposition in irreducible factors:

u^5-4u^4+14u^3-20u^2+25u = u (u^2-2u+5)^2

We now try to find A, B, C, D, E \in \mathbb{R} such that

F(u) = \frac{2u^4-5u^3+24u^2-27u+50}{u (u^2-2u+5)^2} = \frac{A}{u} + \frac{Bu+C}{u^2-2u+5} + \frac{Du+E}{(u^2-2u+5)^2} \lim_{u \rightarrow 0} uF(u) = A = \frac{50}{3 \times 25} = 2 \lim_{u \rightarrow +\infty} uF(u) = A + B = 2 \rightarrow B = 0 \lim_{u \rightarrow +\infty} u^2\left(F(u) - \frac{A}{u}\right) = C = \lim_{u \rightarrow +\infty} u^2 \frac{2 u^4 - 5u^3 - 2(u^4-4u^3)}{u \times u^4} = 3 F(1) = \frac{11}{4} = 2 + \frac{3}{4} + \frac{D+E}{16} \rightarrow D+E = 44 - 32 - 12 = 0 F(2) = \frac{42}{25} = 1 + \frac{3}{5} + \frac{2D+E}{25} \rightarrow 2D + E = 42 - 25 - 15 = 2

From the two last equalities, we immediately get D = 2 and E = -2. Finally,

F(u) = \frac{2}{u} + \frac{3}{u^2-2u+5} + \frac{2u - 2}{(u^2-2u+5)^2}

We have

\int_{u=1}^{5} \frac{2}{u} d u= \left[ 2 \ln|u| \right]_{u=1}^5 = 2 \ln(5) \int_{u=1}^{5} \frac{3}{u^2-2u+5} d u= \frac{3}{4} \int_{u=1}^{5} \frac{d u}{(\frac{u-1}{2})^2+1} = \frac{3}{2} \int_{v=0}^{2} \frac{d v}{v^2+1} = \frac{3}{2} \arctan(2) \int_{u=1}^{5} \frac{2u - 2}{(u^2-2u+5)^2} d u = \left[ - \frac{1}{u^2-2u+5} \right]_{u=1}^5 = \frac{1}{5}

hence

\int_{u=1}^5 F(u) d u = 2 \ln(5) + \frac{3}{2} \arctan(2) + \frac{1}{5}

#### Example 3

\int_{t=2}^{17} \frac{t^6-2t^5-3t^4+15t^3+4t^2-17}{t^4+2t^3-t-2} d t

By polynomial long division we find that

X^6-2x^5-3X^4+15X^3+4X^2-17 = (X^2-4X+5)(X^4+2X^3-X-2)+6X^3+2X^2-3X-7

and so

\frac{t^6-2t^5-3t^4+15t^3+4t^2-17}{t^4+2t^3-t-2} = t^2-4t+5+\frac{6t^3+2t^2-3t-7}{t^4+2t^3-t-2}

The integral of the first term is:

\int_{t=2}^{17} t^2-4t+5 d t = {\left[\frac{t^3}{3} - 2t^2 + 5t\right]}_{t=2}^{17} = 1140

First we search trivial roots for the denominator. If we evaluate t^4+2t^3-t-2 at t=0,-1,1,-2,2 we find respectively -2, -2, 0, 0, 28. So we can factor X^4+2X^4-X-2 by (X - 1)(X + 2). We get

X^4+2X^3-X-2 = (X - 1)(X + 2)(X^2+X+1)

The polynomial X^2+X+1 is irreducible on \mathbb{R} because its discrimant is 1 - 4 = -3 < 0. Hence we search a decomposition of

F(t) = \frac{6t^3+2t^2-3t-7}{(t - 1)(t + 2)(t^2+t+1)} = \frac{A}{t - 1} + \frac{B}{t + 2} + \frac{C t + D}{t^2+t+1}

for some constants A, B, C, D \in \mathbb{R}. We have

\lim_{t \rightarrow 1} (t - 1)F(t) = A = \frac{6+2-3-7}{3 \times 3} = - \frac{2}{9} \lim_{t \rightarrow -2} (t + 2)F(t) = B = \frac{6\times-8+2\times4+3\times2-7}{-3 \times (4-2+1)} = \frac{41}{9} \lim_{t \rightarrow +\infty} t F(t) = A + B + C = 6 \rightarrow C = 6 + \frac{2}{9} - \frac{41}{9} = \frac{5}{3} F(0) = -A + \frac{B}{2} + D = \frac{-7}{-1 \times 2 \times 1} \rightarrow D = \frac{7}{2} -\frac{2}{9} - \frac{41}{18} = 1

So

F(t) = -\frac{2}{9(t - 1)} + \frac{41}{9(t + 2)} + \frac{\frac{5}{3} t + 1}{t^2+t+1}

The integral of the two first terms is easy to compute:

\int_{t=2}^{17} -\frac{2}{9(t - 1)} + \frac{41}{9(t + 2)} d t = -\frac{2}{9} \left[\ln|t-1|\right]_{2}^{17} + \frac{41}{9} \left[\ln|t+2|\right]_{2}^{17} = -\frac{2}{9} \ln(16) + \frac{41}{9} \ln(19) - \frac{41}{9} \ln(4)

The last term can be split into two parts:

\frac{\frac{5}{3} t + 1}{t^2+t+1} = \frac{5}{6} \frac{2t+1}{t^2+t+1} + \frac{1}{6} \times \frac{4}{3} \frac{1}{\left( \frac{t+\frac{1}{2}}{\sqrt{\frac{3}{4}}} \right)^2 + 1}

We have

\int_{t=2}^{17} \frac{5}{6} \frac{2t}{t^2+t+1} d t = \frac{5}{6} \left[ \ln |t^2 + t + 1| \right]_{t=2}^{17} = \frac{5}{6} \ln(307) - \frac{5}{6} \log(7)

and

\int_{t=2}^{17} \frac{2}{9} \frac{d t}{\left( \frac{t+\frac{1}{2}}{\sqrt{\frac{3}{4}}} \right)^2 + 1} = \frac{2}{9} \int_{u=\frac{5}{\sqrt{3}}}^{\frac{35}{\sqrt{3}}} \frac{\sqrt{\frac{3}{4}} d u}{u^2 + 1} = \frac{1}{3 \sqrt{3}} \left[ \arctan u \right]_{u=\frac{5}{\sqrt{3}}}^{\frac{35}{\sqrt{3}}}

Finally we get

\int_{t=2}^{17} \frac{t^6-2t^5-3t^4+15t^3+4t^2-17}{t^4+2t^3-t-2} d t = 1140 -\frac{2}{9} \ln(16) + \frac{41}{9} \ln(19) - \frac{41}{9} \ln(4) + \frac{5}{6} \ln(307) - \frac{5}{6} \log(7) + \frac{1}{3 \sqrt{3}} \arctan\left(\frac{35}{\sqrt{3}}\right) - \frac{1}{3 \sqrt{3}} \arctan\left(\frac{5}{\sqrt{3}}\right)

Alternatively, we can write

G(t) = \frac{\frac{5}{3} t + 1}{t^2+t+1} = \frac{E}{t - t^+} + \frac{F}{t - t^-}

where t^\pm = \frac{-1\pm i \sqrt{3}}{2} are the complex roots of t^2+t+1. We have

\lim_{t \rightarrow t^+} (t-t^+) G(t) = E = \frac{5\sqrt{3}+i}{6 \sqrt{3}} \lim_{t \rightarrow t^-} (t-t^-) G(t) = F = \frac{5\sqrt{3}-i}{6 \sqrt{3}}

We have

\int_{2}^{17} G(t) d t = \left[ E \left( \ln |t - t^+| + i \arctan \frac{t - \Re(t^+)}{\Im(t^+)} \right) + F \left( \ln |t - t^-| + i \arctan \frac{t - \Re(t^-)}{\Im(t^-)} \right) \right]_{2}^{17}

After simplification, we get

\int_{2}^{17} G(t) d t = \frac{5}{6} \ln(307) - \frac{5}{6} \log(7) + \frac{1}{3 \sqrt{3}} \arctan\left(\frac{35}{\sqrt{3}}\right) - \frac{1}{3 \sqrt{3}} \arctan\left(\frac{5}{\sqrt{3}}\right)

### Contour Integration

#### Idea

Consider contour integral \oint_\gamma f(z) d z along paths in complex plane to deduce integral along the real line.

#### Example 1

Consider the integral

\int_{0}^{2 \pi} \frac{6 \cos(2 \theta)}{4 \sin(\theta) - 5} d\theta

Write z = e^{i \theta}, d z = i z d \theta. The integral becomes

\oint_{\gamma} \frac{3 (z^2 + z^{-2})}{-2 i (z - z^{-1}) - 5} \frac{d z}{i z} = \oint_{\gamma} \frac{3(z^4+1)}{2 z^2 (z - 2 i)\left(z - \frac{i}{2}\right)} d z

where \gamma is the unit circle traversed counterclockwise. f : z \mapsto \frac{3(z^4+1)}{2 z^2 (z - 2 i)\left(z - \frac{i}{2}\right)} has two singularities inside that circle: 0 and \frac{i}{2}. Hence the integral is the 2 \pi i times the sum of the residues of f at these points:

\mathrm{Res}(f, 0) = \lim_{z \rightarrow 0} \frac{d \left( z^2 f(z) \right)}{dz} = \frac{15 i}{4} \mathrm{Res}(f, \frac{i}{2}) = \lim_{z \rightarrow 0} \left( z - \frac{i}{2} \right) f(z) = -\frac{17 i}{4}

Hence

\int_{0}^{2 \pi} \frac{6 \cos(2 \theta)}{4 \sin(\theta) - 5} d\theta = \oint_{\gamma} f(z) d z = 2 \pi i \left( \frac{15 i}{4} -\frac{17 i}{4} \right) = \pi

#### Example 2

We want to know the value of the integral

\int_{-\infty}^{+\infty} \frac{16 \sin(2x+\pi)}{5(x^2+2x+5)^2} dx

For that purpose, we introduce the function

f(z) = \frac{16 e^{-2\pi z}}{5(z^2+2z+5)^2}

We note that x \mapsto \Im(f(x)) is the integrand above. We also note that the poles of f are z^\pm = -1 \pm 2 i. By choosing the right contour, one can show that

\int_{-\infty}^{+\infty} f(x) dx = -2 i \pi \sum_{\stackrel{z\text{ residue}}{\Im(z) < 0}} \mathrm{Res}(f, z)

We have

\mathrm{Res}(f, z^-) = \lim_{z \rightarrow z⁻} \frac{d \left( (z - z^-)^2 f(z) \right)}{dz} = \frac{i e^{2 i}}{2 e^4}

Hence

\int_{-\infty}^{+\infty} \frac{16 \sin(2x+\pi)}{5(x^2+2x+5)^2} dx = \Im\left(\int_{-\infty}^{+\infty} f(x) dx\right) = \Im\left(-2 i \pi \frac{i e^{2 i}}{2 e^4}\right) = \Im\left(\frac{\pi e^{2 i}}{e^4}\right)

Finally

\int_{-\infty}^{+\infty} \frac{16 \sin(2x+\pi)}{5(x^2+2x+5)^2} dx = \frac{\pi \sin(2)}{e^4}