Integrals

1. Methods of Computation
- 1.1. Method of Substitution
- 1.2. Integration by Parts
2. Tricks
- 2.1. Weierstrass Substitution
3. References
- 3.1. Treatises

1. Methods of Computation

The only way we can do integration analytically is from the fundamental theorem of calculus.

Let \(f\colon[a,b]\to\RR\), and \(F\colon[a,b]\to\RR\) differentiable such that \(F'(x)=f(x)\) for each \(x\in[a,b]\). Then

\begin{equation} \int^{b}_{a}f(x)\,\D x = F(b)-F(a). \end{equation}

Let \(f\colon[a,b]\to\RR\) be continuous and \(F\) its antiderivative. Then for any \(x\in[a,b]\), we have

\begin{equation} \frac{\D}{\D x}\int^{x}_{a}f(t)\,\D t = f(x). \end{equation}

Now the two main methods of computing integrals may be deduced immediately.

1.1. Method of Substitution

Let \(\varphi\) be differentiable, \(f\) some function. Then

\begin{equation} \int f(\varphi(t))\varphi'(t)\,\D t = \int f(u)\,\D u. \end{equation}

Definite integration similarly is defined

\begin{equation} \int^{b}_{a} f(\varphi(t))\varphi'(t)\,\D t = \int^{\varphi(b)}_{\varphi(a)} f(u)\,\D u. \end{equation}

The heuristic is \(\D u = \varphi'(t)\,\D t\), and the boundary of integration is determined by \(\varphi([a,b])\).

1.2. Integration by Parts

Let \(f,g\colon[a,b]\to\RR\) be such that \(g\) is differentiable on \([a,b]\) whose first derivative is integrable, and \(f\) differentiable on \([a,b]\) (possibly with at most countably many discontinuities), then

\begin{equation} \int^{b}_{a}f(x)g'(x)\,\D x = f(b)g(b)-f(a)g(a)-\int^{b}_{a}f'(x)g(x)\,\D x. \end{equation}

This may be obtained by the fundamental theorem of calculus applied to the product rule:

\begin{equation} \int^{b}_{a}\frac{\D}{\D x}\bigl(f(x)g(x)\bigr)\D x = f(b)g(b)-f(a)g(a) \end{equation}

and expanding the left-hand side using the product rule

\begin{equation} \int^{b}_{a}\frac{\D}{\D x}\bigl(f(x)g(x)\bigr)\D x = \int^{b}_{a}f'(x)g(x)\,\D x + \int^{b}_{a}f(x)g'(x)\,\D x. \end{equation}

Setting these two equal give us

\begin{equation} \int^{b}_{a}f'(x)g(x)\,\D x + \int^{b}_{a}f(x)g'(x)\,\D x = f(b)g(b)-f(a)g(a). \end{equation}

The result follows immediately.

Let \(f,g\colon\RR\to\RR\) be "sufficiently nice" functions. Then for indefinite integrals, we have

\begin{equation} \int f(x)g'(x)\,\D x = f(x)g(x) - \int f'(x)g(x)\,\D x. \end{equation}

2. Tricks

Let \(a>0\) and \(f\colon[-a,a]\to\RR\) be continuous. If \(f\) is even, then

\begin{equation} \int^{a}_{-a}\frac{f(x)}{1 + \E^{x}}\D x = \int^{a}_{0}f(x)\,\D x. \end{equation}

(I first saw this on math.SX)

Only the even part survives, so

\begin{equation} h(x) = \frac{1}{2}\left(\frac{f(x)}{1 + \E^{x}} + \frac{f(-x)}{1 + \E^{-x}}\right) \end{equation}

But we see

\begin{equation} \frac{1}{1+\E^{x}} + \frac{1}{1+\E^{-x}} = \frac{2 + \E^{x} + \E^{-x}}{1 + \E^{x} + \E^{-x} + 1} = 1 \end{equation}

hence

\begin{equation} h(x) = \frac{1}{2}(f(x) + f(-x)) = f(x). \end{equation}

So the integral becomes

\begin{equation} 2\int^{a}_{0}h(x)\,\D x = 2\int^{a}_{0}f(x)\,\D x \end{equation}

as desired.

There was nothing special about using \((1 + \E^{x})\), we could use any \((1 + c^{x})\) for positive real number \(c\gt0\).

2.1. Weierstrass Substitution

Take \(t = \tan(\theta/2)\), then

\begin{equation} \sin(\theta)=\frac{2t}{1+t^{2}},\quad\cos(\theta)=\frac{1-t^{2}}{1+t^{2}} \end{equation}

and in particular

\begin{equation} \D\theta = \frac{2\,\D t}{1 + t^{2}}. \end{equation}

3. References

Michael Dougherty, Advanced Integration Techniques, 76 pages.
Victor Moll's papers about the proofs of Gradshteyn and Ryzhik
Keith Conrad, The Gamma Function.

3.1. Treatises

Joseph Edward, A treatise on the integral calculus; with applications. vol 1, vol 2 1921–1922.