change of variable | College Math Teaching

September 13, 2021

Integrals of functions with nice inverses

Filed under: change of variable, derivatives, elementary mathematics, integrals, integration by substitution — collegemathteaching @ 4:49 pm

This idea started as a bit of a joke:

https://twitter.com/lightediand/status/1437181306526371845

Of course, for readers of this blog: easy-peasy. $u =\sqrt{tan(x)} \rightarrow u^2 =tan(x) \rightarrow x = arctan(u^2), dx = {2udu \over 1+u^4}$ so the integral is transformed into $\int {2u^2 \over 1+u^4} du$ and so we’ve entered the realm of rational functions. Ok, ok, there is some work to do.

But for now, notice what is really doing on: we have a function under the radical that has an inverse function (IF we are careful about domains) and said inverse function has a derivative which is a rational function

More shortly: let $f(x)$ be such that ${d \over dx} f^{-1}(x) = q(x)$ then:

$\int (f(x))^{1 \over n} dx$ gets transformed: $u^n = f(x) \rightarrow x =f^{-1}(u^n)$ and then $dx = nu^{n-1}q(u^n)$ and the integral becomes $\int n u^n q(u^n) du$ which is a rational function integral.

Yes, yes, we need to mind domains.

July 14, 2020

An alternative to trig substitution, sort of..

Filed under: calculus, change of variable, derivatives, integrals, integration by substitution — Tags: hyperbolic trigonometric functions — oldgote @ 1:52 am

Ok, just for fun: $\int \sqrt{1+x^2} dx =$

The usual is to use $x =tan(t), dx =sec^2(t) dt$ which transforms this to the dreaded $\int sec^3(t) dt$ integral, which is a double integration by parts.
Is there a way out? I think so, though the price one pays is a trickier conversion back to x.

Let’s try $x =sinh(t) \rightarrow dx = cosh(t) dt$ so upon substituting we obtain $\int |cosh(t)|cosh(t) dt$ and noting that $cosh(t) > 0$ alaways:

$\int cosh^2(t)dt$ Now this can be integrated by parts: let $u=cosh(t) dv = cosh(t) dt \rightarrow du =sinh(t), v = sinh(t)$

So $\int cosh^2(t)dt = cosh(t)sinh(t) -\int sinh^2(t)dt$ but this easily reduces to:

$\int cosh^2(t)dt = cosh(t)sinh(t) -\int cosh^2(t)-1 dt \rightarrow 2\int cosh^2(t)dt = cosh(t)sinh(t) -t + C$

Division by 2: $\int cosh^2(t)dt = \frac{1}{2}(cosh(t)sinh(t)-t)+C$

That was easy enough.

But we now have the conversion to x: $\frac{1}{2}(cosh(t)sinh(t) \rightarrow \frac{1}{2}x \sqrt{1+x^2}$

So far, so good. But what about $t \rightarrow arcsinh(x)$ ?

Write: $sinh(t) = \frac{e^{t}-e^{-t}}{2} = x \rightarrow e^{t}-e^{-t} =2x \rightarrow e^{t}-2x -e^{-t} =0$

Now multiply both sides by $e^{t}$ to get $e^{2t}-2xe^t -1 =0$ and use the quadratic formula to get $e^t = \frac{1}{2}(2x\pm \sqrt{4x^2+4} \rightarrow e^t = x \pm \sqrt{x^2+1}$

We need $e^t > 0$ so $e^t = x + \sqrt{x^2+1} \rightarrow t = ln|x + \sqrt{x^2+1}|$ and that is our integral:

$\int \sqrt{1+x^2} dx = \frac{1}{2}x \sqrt{1+x^2} + \frac{1}{2} ln|x + \sqrt{x^2+1}| + C$

I guess that this isn’t that much easier after all.

Comments (2)

March 16, 2019

The beta function integral: how to evaluate them

Filed under: analysis, calculus, change of variable, density function, improper integrals, integrals, integration by substitution, likelihood function, statistics — Tags: beta distribution, beta function, change of variables, gamma function, Jacobian — collegemathteaching @ 10:30 pm

My interest in “beta” functions comes from their utility in Bayesian statistics. A nice 78 minute introduction to Bayesian statistics and how the beta distribution is used can be found here; you need to understand basic mathematical statistics concepts such as “joint density”, “marginal density”, “Bayes’ Rule” and “likelihood function” to follow the youtube lecture. To follow this post, one should know the standard “3 semesters” of calculus and know what the gamma function is (the extension of the factorial function to the real numbers); previous exposure to the standard “polar coordinates” proof that $\int^{\infty}_{-\infty} e^{x^2} dx = \sqrt{\pi}$ would be very helpful.

So, what it the beta function? it is $\beta(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$ where $\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t} dt$ . Note that $\Gamma(n+1) = n!$ for integers $n$ The gamma function is the unique “logarithmically convex” extension of the factorial function to the real line, where “logarithmically convex” means that the logarithm of the function is convex; that is, the second derivative of the log of the function is positive. Roughly speaking, this means that the function exhibits growth behavior similar to (or “greater”) than $e^{x^2}$

Now it turns out that the beta density function is defined as follows: $\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1}(1-x)^{b-1}$ for $0 < x < 1$ as one can see that the integral is either proper or a convergent improper integral for $0 < a < 1, 0 < b < 1$ .

I'll do this in two steps. Step one will convert the beta integral into an integral involving powers of sine and cosine. Step two will be to write $\Gamma(a) \Gamma(b)$ as a product of two integrals, do a change of variables and convert to an improper integral on the first quadrant. Then I'll convert to polar coordinates to show that this integral is equal to $\Gamma(a+b) \beta(a,b)$

Step one: converting the beta integral to a sine/cosine integral. Limit $t \in [0, \frac{\pi}{2}]$ and then do the substitution $x = sin^2(t), dx = 2 sin(t)cos(t) dt$ . Then the beta integral becomes: $\int_0^1 x^{a-1}(1-x)^{b-1} dx = 2\int_0^{\frac{\pi}{2}} (sin^2(t))^{a-1}(1-sin^2(t))^{b-1} sin(t)cos(t)dt = 2\int_0^{\frac{\pi}{2}} (sin(t))^{2a-1}(cos(t))^{2b-1} dt$

Step two: transforming the product of two gamma functions into a double integral and evaluating using polar coordinates.

Write $\Gamma(a) \Gamma(b) = \int_0^{\infty} x^{a-1} e^{-x} dx \int_0^{\infty} y^{b-1} e^{-y} dy$

Now do the conversion $x = u^2, dx = 2udu, y = v^2, dy = 2vdv$ to obtain:

$\int_0^{\infty} 2u^{2a-1} e^{-u^2} du \int_0^{\infty} 2v^{2b-1} e^{-v^2} dv$ (there is a tiny amount of algebra involved)

From which we now obtain

$4\int^{\infty}_0 \int^{\infty}_0 u^{2a-1}v^{2b-1} e^{-(u^2+v^2)} dudv$

Now we switch to polar coordinates, remembering the $rdrd\theta$ that comes from evaluating the Jacobian of $x = rcos(\theta), y = rsin(\theta)$

$4 \int^{\frac{\pi}{2}}_0 \int^{\infty}_0 r^{2a +2b -1} (cos(\theta))^{2a-1}(sin(\theta))^{2b-1} e^{-r^2} dr d\theta$

This splits into two integrals:

$2 \int^{\frac{\pi}{2}}_0 (cos(\theta))^{2a-1}(sin(\theta))^{2b-1} d \theta 2\int^{\infty}_0 r^{2a +2b -1}e^{-r^2} dr$

The first of these integrals is just $\beta(a,b)$ so now we have:

$\Gamma(a) \Gamma(b) = \beta(a,b) 2\int^{\infty}_0 r^{2a +2b -1}e^{-r^2} dr$

The second integral: we just use $r^2 = x \rightarrow 2rdr = dx \rightarrow \frac{1}{2}\frac{1}{\sqrt{x}}dx = dr$ to obtain:

$2\int^{\infty}_0 r^{2a +2b -1}e^{-r^2} dr = \int^{\infty}_0 x^{a+b-\frac{1}{2}} e^{-x} \frac{1}{\sqrt{x}}dx = \int^{\infty}_0 x^{a+b-1} e^{-x} dx =\Gamma(a+b)$ (yes, I cancelled the 2 with the 1/2)

And so the result follows.

That seems complicated for a simple little integral, doesn’t it?

Comments (5)

February 19, 2012

Divergent Improper Integrals: change of variables to an unbounded integrand.

Filed under: calculus, change of variable, improper integrals, integrals, integration by substitution, pedagogy — collegemathteaching @ 10:41 pm

This post was motivated by a student question: my student wanted help with the following problem:

$\int^{\infty}_{1} \frac{x^2}{\sqrt{x^3 +1}} dx$
Of course the idea is to do a substitution: $u = x^3 + 1$ which transforms the integral into $\frac{1}{3} \int^{\infty}_{2} \frac{1}{\sqrt{u}} du$ which diverges. So far, so good. But then I told him one of my calculus tips: “it is often a good idea to try to guess the answer ahead of time” and then pointed out that for large values of $x, \frac{x^2}{\sqrt{x^3 +1}} \approx \frac{x^2}{\sqrt{x^3}} = \sqrt{x}$ and of course $\int^{\infty}_1 \sqrt{x} dx$ diverges because the integrand does not go to zero (in fact, is unbounded!) as $x$ tends to infinity.

Then I realized that a change of variables had taken an unbounded function to a bounded one…though one which did not produce a convergent improper integral.

That lead to the natural question: if one has an integrand which is positive but monotonically decreasing to zero on $[1, \infty )$ , is there a change of variables which will change the integrand to either an unbounded function on $[1, \infty )$ or at least one that does not decrease to zero?

I admit that I have not answered this question yet, nor have I looked it up. But I can answer this question for a certain class of functions:

Theorem

Given $\int^{\infty}_1 \frac{1}{x^r} dx$
If $0 < r < 1$ , let $k > \frac{1}{1-r}$ . Then the change of variable $u = x^{\frac{1}{k}}$ transforms $\int^{\infty}_{1} \frac{1}{x^r} dx$ to $k \int^{\infty}_1 u^{k(1-r) -1} du$ and of course $u^{k(1-r) -1}$ is unbounded on $[1, \infty)$ .

If $1 < r$ let $k < \frac{1}{1-r} < 0$ . Then $\int^{\infty}_1 \frac{1}{x^r} dx$ is transformed into $|k|\int^{1}_{0} u^{-1+k(1-r)} du$ which is an integral of a bounded function over a bounded region.

In short, one class of functions whose improper integral diverges can be transformed to functions that tend to infinity and the class of functions whose integrals converge can be transformed into functions which are bounded over a bounded interval.

Here is such an example: We show the equivalent integrals $\int^{1.5}_{1} 3x^{\frac{1}{2}} dx$ and $\int^{(1.5)^3}_1 \frac{1}{\sqrt{u}} du$ . The transformation is accomplished by using $u =x^3$ . Note how the transformation stretches the interval of integration to account for the function “shrinkage”.