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

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

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 $\int_0^\infty e^{-t^2} \space dt$.)

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.)

-
Just to add a bit to Peter's answer, although you cannot solve for the antiderivative explicitly, it is not at all hard to write down a power series for it (just integrate the power series for $e^{-t^2}$). You can get very good estimates on particular definite integrals using that power series expansion and basic techniques on estimating infinite sums. –  Chris Janjigian Apr 30 '12 at 0:48
This integral has no elementary anti-derivative. However, you can figure out $\int_{-\infty}^{\infty} e^{-x^2} \,dx$ by squaring the integral and turning it into polar coordinates. –  Argon Apr 30 '12 at 0:49
Nice suggestions Chris and Argon, thank you. –  Joe Apr 30 '12 at 0:56
Any reason for the downvote? –  Joe Jun 26 '12 at 14:54
add comment

## 3 Answers

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.

-
+1. Lovely answer. Many thanks Peter. –  Joe Apr 30 '12 at 0:56
If need be, Liouville's theorem can be used to show that the Gaussian function's antiderivative cannot be expressed elementarily. –  Ｊ. Ｍ. Apr 30 '12 at 0:59
@J.M. That seems rather too advanced. –  Pedro Tamaroff Apr 30 '12 at 1:00
As I said, "if need be". If somebody isn't convinced by the fact that we have to define the error function, that's where you can bring up Liouville's theorem. –  Ｊ. Ｍ. Apr 30 '12 at 1:02
@J.M. Oh. Sorry, I didn't read the "If need be" part. You're right. –  Pedro Tamaroff Apr 30 '12 at 1:10
add comment

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.

-
add comment

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!

-
add comment