How to integrate $\int e^{-t^{2}} \space \, \mathrm dt $ using introductory calculus methods

Question

Earlier today I stumbled across this when I was doing some practice questions for a physics course: $$\int e^{-t^2} \space \, \mathrm dt $$

To expand, the limits of integration were something like $1$ and $4$ (it was just a velocity function that needed to be integrated to find distance - it was not a known integral like $\int_0^\infty e^{-t^2} \space\, \mathrm dt$.)

Based on Wolfram|Alpha, it appears it cannot be expressed in elementary terms (i.e. it involves the error function.) Note that the questions involved the use of a calculator, so I was able to integrate the function using a CAS with ease, but I am wondering how to do it by hand. Thus, I was wondering if there was possibly a way to evaluate the integral using elementary methods from a calculus one or two course (read: no complex analysis). I thought there may perhaps be an elementary solution (I don't know what kind of algorithm Wolfram uses to evaluate integrals - I have seen them evaluate easy integrals in a lot of steps before.)

Answer 1

The error function is defined as

$$\operatorname{erf}(x)=\frac 2 {\sqrt \pi}\int_{0}^x e^{-t^2}dt$$

It is not an elementary function. Since from the definition it is immediate (FTCI) that

$$\operatorname{erf}'(x)=\frac 2 {\sqrt \pi}e^{-x^2}$$

the primitive of $e^{-x^2}$ is expressible as

$$\int e^{-x^2} dx =\frac{\sqrt \pi}{2}\operatorname{erf}(x)-C $$

since any two primitives of a function $f$ differ by a constant (FTCII)

As a consequence your primitive can't be expressed in terms of elementary functions.

Answer 2

Your only hope is to spread out the first few terms of its Taylor series. Just substitute $-t^2$ into the exponential series $1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} +\frac{x^4}{4!}$ as far out as you can stand. When you make your substitution, you'll have an alternating sum which has some special properties to help you evaluate your error...and there will be error. There's no getting around it unless as you said you have convenient limits like $0$ to $+\infty$ which you already mentioned. Good Luck! This is just a less strenuous route which conveniently avoids using erf.

Answer 3

As others have said, there is no integrated form in elementary functions.

You could use the trapezium rule (T), or Simpson's rule (S);

or you could use Romberg integration: Start with (T), with a suitable number of subintervals, e.g., 5, 10, 20 -- in this case, 15 would help, because of the length of the interval. Call the result T11.

Double the number of strips, and calculate T21 in the same way. (You need only calculate the in-between ordinates!)

Then let T22 = (4*T21 - T11) / 3. Double again, to get T31; then T32 = (4*T31 - T21) / 3;
and T33 = (4^2 * T32 - T22) / (4^2 - 1) Continue in this way until Tnn = t(n-1)(n-1).

Try it in some simple form to check that it works -- e.g., ∫ 1/(1 + x^2) = arctan x + c

Note that it should work exactly for a quadratic expression at T22, (and for a cubic) because the second level is Simpson's rule.

Use a spreadsheet for the calculations!

Answer 4

While integrating the power series of $ e^{-x^2} $ is the easiest option, but it requires computation of sum of more number of initial terms of the resulting infinite series to converge to a good accuracy result. I feel that the better convergence is obtained in lesser number of terms by using integration by parts technique as I had shown below.

We have, $\int e^{-x^2}dx=xe^{-x^2}+2\int x^2e^{-x^2}dx$

$\int e^{-x^2}dx=xe^{-x^2}+\frac{2}{3}x^3e^{-x^2}+\frac{4}{3}\int x^4e^{-x^2}dx$

Continuing like this, we get a following infinite series, $\int e^{-x^2}dx=e^{-x^2} (x+\frac{2x^3}{1\times 3}+\frac{2^2x^5}{3\times 5}+\frac{2^3x^7}{3\times 5\times 7}+......)$ $\int e^{-x^2}dx=e^{-x^2} \sum_{r=1}^{\infty}\frac{2^{r-1}x^{2r-1}}{\prod_{k=1}^{r}(2k-1)}$

Using Mathematica, I found that even for evaluating the above integral at comparatively smaller range of interval such as 0 to 10 i.e. $ \int_{0}^{10} e^{-x^2}dx $, computing sum of first 280 terms of power series integral of $e^{-x^2}$ gives result as $0.886227...$ but this result is achieved by computing sum of first only 153 terms of the series I suggested above.(I leave this part to you to explore more).

Also, if the range of integral is within 0 to 1, then almost same accuracy in the result can be expected if we evaluate the integral using both the ways by considering the sum of same number of initial terms.(I have verified this)

So, the series I suggested is a bit quicker than the simple power series when the range of integral is more than 0 to 1.