The foundation of a technique for evaluating definite integrals |

Given $f(x)$ is continuous on $[a, b]$ and $\phi(t)$ satisfies the following conditions:

[1] $\forall t \in [\alpha, \beta], \phi(t) \in [a, b]$

[2] $\phi(\alpha)=a, \; \phi(\beta)=b$

[3] $\phi(t)$ has continuous derivative on $[\alpha, \beta]$

Prove:

$\displaystyle\int\limits_{a}^{b}f(x)\;dx =\displaystyle\int\limits_{\alpha}^{\beta}f(\phi(t))\phi'(t)\;dt.$

The given premises

$f(x)$ is a continuous function on $[a, b]\quad\quad\quad(1)$

ensures the existence of the definite integral $\displaystyle \int\limits_{a}^{b}f(x)\;dx$ and the antiderivative of $f(x).$

Denoting the antiderivative of $f(x)$ as $F(x),$ we obtain

$\displaystyle\int\limits_{a}^{b} f(x) \;dx = F(b)-F(a).\quad\quad\quad(2)$

We also deduce from [3] that

$\phi(t)$ is continuous on $[\alpha, \beta].\quad\quad\quad(3)$

Combining [3], [1] and (1),

$f(\phi(t))$ is continuous on $[\alpha, \beta].\quad\quad\quad(4)$

Additionally, as per [3],

$\phi'(t)$ is continuous on $[\alpha, \beta].\quad\quad\quad(5)$

Together, (4) and (5) give

$f(\phi(t))\phi'(t)$ is continuous on $[\alpha, \beta].$

Consequently, $\displaystyle\int\limits_{\alpha}^{\beta}f(\phi(t))\phi'(t)\;dt$ exists as well.

Now let’s examine $F(u)$ , where $u=\phi(t)$ . We have

$\frac{dF(u)}{dx} = F'(u) \frac{du}{dx} = F'(u)\phi'(t) = f(u)g'(t) = f(\phi(t))\phi'(t)$

$\implies F(u) = f(\phi(t))$ is the antiderivative of $f(\phi(t))\phi'(t)$

$\implies \displaystyle\int\limits_{\alpha}^{\beta}f(\phi(t))\phi'(t)\;dt = F(\phi(t))\bigg|_{\alpha}^{\beta}$

$= F(\phi(\beta))-F(\phi(\alpha)) \overset{[2]}{=} F(b)-F(a)\overset{(2)}{=}\displaystyle\int\limits_{a}^{b}f(x)\;dx.$

This theorem serves as the foundation of a technique widely employed in evaluating definite integrals. By selecting a suitable substitution, the definite integral is transformed into a form that is more manageable or matches a known solution. This method proves particularly valuable for evaluating definite integrals involving complex functions or expressions.

For Example, to evaluate integral

$\displaystyle\int\limits_{0}^{a}\log(x+\sqrt{x^2+a^2})\;dx \quad (a>0)$ ,

we choose the following substitution for variable $x$ :

$\phi(t) = at \implies \phi'(t) =a.$

Since $0 \le \phi(t) \le a \implies 0 \le t \le 1,$ we can rewrite the integral as:

$\displaystyle\int\limits_{0}^{1}\left(\log(at + \sqrt{(at)^2+a^2}\right)\cdot a \; dt$

and proceed to evaluate the new integral as follows:

$\displaystyle\int\limits_{0}^{1}\left(\log(at + |a|\sqrt{t^2+1}\right)\cdot a \; dt$

$\overset{a>0}{=} \displaystyle\int\limits_{0}^{1}\left(\log(at + a\sqrt{t^2+1}\right)\cdot a \; dt$

$= \displaystyle\int\limits_{0}^{1}a\left(\log(a\cdot(t+\sqrt{t^2+1})\right)\; dt$

$= \displaystyle\int\limits_{0}^{1}a\left(\log(a)+\log(t+\sqrt{t^2+1})\right)\; dt$

$= \displaystyle\int\limits_{0}^{1}a\log(a)\;dt+a\int\limits_{0}^{1}\log(t+\sqrt{t^2+1})\; dt$

$= \displaystyle a\log(a) t\bigg|_{0}^{1} + a\int\limits_{0}^{1}\mathrm{arcsinh(t)}\;dt$

$= a\log(a) + a\cdot\left(t\cdot\mathrm{arcsinh(t)} - \sqrt{t^2+1}\right)\bigg|_{0}^{1}$

$= a\log(a) + a\cdot\left(\mathrm{arcsinh(1)}-\sqrt{2} + 1\right)$

$= a\log(a) + a\cdot\mathrm{arcsinh(1)} -\sqrt{2}a + a.$

Let’s verify the result:

Suppose $G(x) = \displaystyle\int \log(x+\sqrt{x^2+a^2})\; dx$ . Then

$G(x) = \displaystyle \int x' \cdot\log(x+\sqrt{x^2+a^2})\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2})-\int x\cdot \frac{1+\frac{1}{2}\frac{2x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}}\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2})-\int x \cdot \frac{(1+\frac{x}{\sqrt{x^2+a^2}})(\sqrt{x^2+a^2}-x)}{a^2}\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2}) -\int x \cdot\frac{\sqrt{x^2+a^2}+x-x-\frac{x^2}{\sqrt{x^2+a^2}}}{a^2}\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2}) - \int x\cdot\frac{\sqrt{x^2+a^2}-\frac{x^2}{x^2+a^2}}{a^2}\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2}) - \int x\cdot\frac{\frac{x^2+a^2-x^2}{\sqrt{x^2+a^2}}}{a^2}\;dx$

$= \displaystyle x\log(x+\sqrt{x^2+a^2}) -\int \frac{x}{\sqrt{x^2+a^2}}\; dx$

$= x\log(x+\sqrt{x^2+a^2}) - \sqrt{x^2+a^2}.$

By Newton-Leibniz rule,

$\displaystyle\int\limits_{0}^{a} \log(x+\sqrt{x^2+a^2}) \; dx=G(a) - G(0)$

$= \underbrace{a\log(a+\sqrt{a^2+a^2})-\sqrt{a^2+a^2}}_{G(a)}-\underbrace{(0 - \sqrt{0+a^2})}_{G(0)}$

$= a\log(a\cdot(1 + \sqrt{1+1}))-\sqrt{2}a-(0 - a)$

$= a\log(a) + a\cdot\log(1+\sqrt{1 + 1}) - \sqrt{2}a + a$

$= a\log(a) + a\cdot\mathrm{arcsinh(1)} -\sqrt{2}a + a.$

For grins, we also verify as follows:

$I(\beta) = \displaystyle\int\limits_{0}^{a}\log(x+\sqrt{x^2+\beta^2})\;dx, \quad\beta>0$

$\implies \displaystyle\frac{dI(\beta)}{d\beta} = \int\limits_{0}^{a} \frac{\partial}{\partial \beta}\log(x + \sqrt{x^2+\beta^2})\;dx$

$= \displaystyle\int\limits_{0}^{a}\frac{1}{x+\sqrt{x^2+\beta^2}}\cdot\frac{1}{2\sqrt{x^2+\beta^2}}\cdot 2\beta\; dx$

$= \displaystyle\int\limits_{0}^{a}\frac{\sqrt{x^2+\beta^2}-x}{\beta^2}\cdot\frac{\beta}{\sqrt{x^2+\beta^2}}\;dx$

$= \displaystyle\int\limits_{0}^{a}\frac{\sqrt{x^2+\beta^2}-x}{\beta}\cdot\frac{1}{\sqrt{x^2+\beta^2}}\;dx$

$= \displaystyle\int\limits_{0}^{a}\frac{1}{\beta}(1-\frac{x}{\sqrt{x^2+a^2}})\;dx$

$= \frac{1}{\beta}(x-\sqrt{x^2+\beta^2})\bigg|_{0}^{a}$

$= \frac{1}{\beta}(a-\sqrt{a^2+\beta^2}+\beta).$

That is,

$\frac{dI(\beta)}{d\beta} = \frac{a}{\beta}+1-\frac{\sqrt{a^2+\beta^2}}{\beta}.$

Integrate it from $1$ to $\beta$ ,

$\displaystyle\int\limits_{1}^{\beta} \frac{dI(\beta)}{d\beta} \;d\beta = \int\limits_{1}^{\beta}\frac{a}{\beta} \;d\beta + \int\limits_{1}^{\beta}1\;d\beta - \int\limits_{1}^{\beta}\frac{\sqrt{\beta^2+a^2}}{\beta}\;d\beta$

$\implies \displaystyle I(\beta) - I(1) = a \log(\beta)\bigg|_{1}^{\beta} + \beta\bigg|_{1}^{\beta} -\int\limits_{1}^{\beta}\frac{\sqrt{\beta^2+a^2}}{\beta}\;d\beta$

$\implies \displaystyle I(\beta) - I(1) = a\log(\beta) + \beta -1 - \int\limits_{1}^{\beta}\frac{\sqrt{\beta^2+a^2}}{\beta}\;d\beta$

$\overset{(*), (**)}{\implies} \displaystyle I(\beta) - \underbrace{(a \cdot \mathrm{arcsinh}(a) - \sqrt{a^2+1}+1)}_{I(1)=\int\limits_{0}^{a}\log(x+\sqrt{x^2+1})\;dx} =$

$a\log(a) + \beta -1 -\underbrace{\left(\sqrt{\beta^2+a^2} - \sqrt{a^2+1} - a \cdot \mathrm{arcsinh}(\frac{a}{\beta})+a\cdot \mathrm{arcsinh}(a)\right)}_{\int\limits_{1}^{\beta}\frac{\sqrt{\beta^2+a^2}}{\beta}\;d\beta}$

$\implies I(\beta) = \mathrm{a\log(\beta) + \beta -\sqrt{\beta^2+a^2} + a\cdot arcsinh(\frac{a}{\beta})}.$

Let $\beta = a,$ we obtain

$\displaystyle\int\limits_{0}^{a}\log(x+\sqrt{x^2+a^2})\; dx = \mathrm{a\log(a) + a -\sqrt{2}a+a\cdot arcsinh(1)} = \mathrm{a\log(a)+a\cdot arcsinh(1) -\sqrt{2}a + a}.$

From “Deriving Two Inverse Functions“, we see that $\log(x+\sqrt{x^2+1}) = \mathrm{arcsinh}(x).$ Therefore,

$I(1) = \displaystyle \int\limits_{0}^{a}\log(x + \sqrt{x^2+1}) \; dx$

$= \displaystyle\int\limits_{0}^{a} \mathrm{arcsinh}(x)\;dx$

$= \displaystyle\int\limits_{0}^{a} x'\cdot \mathrm{arcsinh}(x)\;dx$

$= \displaystyle x\cdot\mathrm{arcsinh}(x)\bigg|_{0}^{a} -\int\limits_{0}^{a}\frac{x}{\sqrt{x^2+1}}\;dx$

$= a\cdot\mathrm{arcsinh}(a) - \sqrt{x^2+1}\bigg|_{0}^{a}$

$= a\cdot \mathrm{arcsinh}(a) - \sqrt{a^2+1}+1.\quad\quad\quad(*)$

Now, letting $\phi(t) = a\cdot t,$ we have

$1 \le at \le \beta \implies \frac{1}{a} \le t \le \frac{\beta}{a}, \phi'(t) = a$ and $\displaystyle \int\limits_{1}^{\beta}\frac{\sqrt{\beta^2+a^2}}{\beta}\;d\beta$ becomes