0
$\begingroup$

Define: $f(x) = e^{-x^2}$, then I define another different function $f(a) = e^a$, where the $a$ domain is related to x by the formula $a =-x^2/2$, but in essence I can consider it as a new function right?

Then I integrate the first function $\int_{-\infty}^{x}e^{-x^2}dx$, but it's not integrable in elementary functions. Then I consider the second function: $\int_{-(-\infty)^2/2}^{-x^2/2} e^a*da$ but this integrable.

Obviously this method is wrong, but I don't understand why. Usually you would do the substitution while you're in the integral, thus $da=g(x)*dx$, but why can't I define a totally new function from the very beginning and then relate it to the function I had?

$\endgroup$
2
  • $\begingroup$ What method? Are you trying to evaluate $\int_{-\infty}^t e^{-x^2} \, dx$ by substitution? $\endgroup$
    – user307169
    Commented Mar 30, 2017 at 19:29
  • $\begingroup$ If you sketch them they represent different areas. e.g. $\int^1_0x^2\space dx \ne \int^1_0 x\space dx$ as it affects the path taken to get from $0$ to $1$. $\endgroup$ Commented Mar 30, 2017 at 19:30

2 Answers 2

2
$\begingroup$

My intent is to give three answers, one which shows evidently that this really is wrong via example, one that explains a way of thinking about what has gone wrong which is illustrative of the issue in general (and might lead one to see how to fix the issue in special cases.) The third is a rigorous approach based on Riemann sums.

A simple example. Let's first make sure this doesn't work: $$\int_0^1 3xdx=x^2\bigg|_0^1=\frac{3}{2}.$$ Moreover, we know that this is "true" as we are simply computing the area of a right triangle with height 3 and width 1. Now imagine we want to 'simplify' by integrating the function $f(y)=y$ with upper limit of integration $3$. Then we get: $$\int_0^3ydy=\frac{y^2}{2}|_0^3=\frac{9}{2}=3\frac{3}{2}.$$ This is exactly the mistake we expected, we get an extra 3.

What's gone wrong in terms of units We can think of what's happened in terms of adjusting units (incorrectly). If $x$ is in yards, then $y$ is naturally the same length in feet. However, in adjusting our limits of integration we have only taken care of one unit conversion, but we are calculating an area, and need to convert our second dimension as well. So instead of the answer from the second integral being in square feet as desired, we get units of "yard feet" which is a silly measure of area.

Why can't it be fixed in general In the preceding example you can fix things by making a unit conversion at the end because feet and yards have a linear relationship. However, if the relationship between $x$ and $y$ varies nonlinearly as $x$ varies, then there is no clear modification to be made which allows you to correct the units at the end. $$\int_a^b e^{-x^2}=\lim_{n\rightarrow\infty}\sum_{i=1}^n e^{-(\frac{b-a}{n}i)^2} \frac{b-a}{n}.$$ On the other hand, $$\int_{a^2}^{b^2} e^{-y}=\lim_{n\rightarrow\infty}\sum_{i=1}^n e^{-(\frac{b^2-a^2}{n}i)} \frac{b^2-a^2}{n}.$$ To equate for these sums would be to account for the differences between $\frac{b-a}{n}$ and $\frac{b^2-a^2}{n}$ in both the exponent and out.

$\endgroup$
2
$\begingroup$

When you change variables in the integrand, you indeed get a new (and, if you choose wisely) easier expression to integrate, but you would no longer be summing over rectangles with equal bases, so you would not be getting the desired area under the curve That is, if $a = -x^2$ you can't just say that $da$ is the same as $dx$; in fact, $$ da = -2x\,dx = -2\sqrt{a}\,dx $$ So the integral would be=come $$ \int_{-\infty}^z e^{-x^2}dx = -\frac12\int_{+\infty}^{-z^2} \frac {e^{a}}{\sqrt{a}}\,da = \frac12\int_{-z^2}^\infty \frac {e^{a}}{\sqrt{a}}\,da $$ And you will indeed find these integrals are equal. But neither can be expressed in terms of elementary functions, unless you take $z-> \infty$.

$\endgroup$
1
  • $\begingroup$ But at the end of the day you should be getting the area of our function, one in a different coordinate system from the original coordinate system. Right? $\endgroup$ Commented Mar 30, 2017 at 23:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .