THE GAUSSIAN INTEGRAL. e x2 dx, and K =

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1 THE GAUSSIAN INTEGRAL KEITH CONRAD Let I e 1 x dx, J e x dx, and K These numbers are positive, and J I/( ) and K I/ π. e πx dx. Theorem. With notation as above, I π, or equivalently J π/, or equivalently K 1. We will give multiple proofs of this result. (Other lists of proofs are in [3] and [8].) The theorem is subtle because there is no simple antiderivative for e 1 x (or e x or e πx ). For comparison, xe 1 x dx can be computed using the antiderivative e 1 x : this integral is First Proof: Polar coordinates The most widely known proof uses multivariable calculus: express J as a double integral and then pass to polar coordinates: J e x dx e y dy e (x +y ) dx dy. This is a double integral over the first quadrant, which we will compute by using polar coordinates. The first quadrant is {(r, θ) : r and θ π/}. Writing x + y as r and dx dy as r dr dθ, J π/ re r dr 1 e r 1 π π 4. e r r dr dθ π π/ Taking square roots, J π/. This method is due to Poisson [8, p. 3].. Second Proof: Another change of variables Our next proof uses another change of variables to compute J, but this will only rely on singlevariable calculus. As before, we have ( ) J e (x +y ) dx dy e (x +y ) dx dy, 1 dθ

2 KEITH CONRAD but instead of using polar coordinates we make a change of variables x yt on the inside integral, with dx y dt, so ( ) ( ) J e y (t +1) y dt dy ye y (t +1) dy dt. Since ye ay dy 1 a for a >, we have J dt (t + 1) 1 π π 4, so J π/. This approach is due to Laplace [6, pp ] and historically precedes the more familiar technique in the first proof above. We will see in the eighth proof that this was not Laplace s first method. For t >, set 3. Third Proof: Differentiating under the integral sign ( t A(t) e dx) x. The integral we want to calculate is A( ) J and then take a square root. Differentiating A(t) with respect to t and using the Fundamental Theorem of Calculus, Let x ty, so A (t) t t e x dx e t e t e x dx. A (t) e t te t y dy te (1+y )t dy. The function under the integral sign is easily antidifferentiated with respect to t: A (t) e (1+y )t t 1 + y dy d dt e (1+y )t 1 + y dy. Letting e t (1+x ) B(t) 1 + x dx, we have A (t) B (t) for all t >, so there is a constant C such that (3.1) A(t) B(t) + C ( for all t >. To find C, we let t + in (3.1). The left side tends to e dx) x while the right side tends to dx/(1 + x ) + C π/4 + C. Thus C π/4, so (3.1) becomes ( t ) e x dx π 4 e t (1+x ) 1 + x dx. Letting t in this equation, we obtain J π/4, so J π/. A comparison of this proof with the first proof is in [19].

3 THE GAUSSIAN INTEGRAL 3 4. Fourth Proof: Another differentiation under the integral sign Here is a second approach to finding J by differentiation under the integral sign. I heard about it from Michael Rozman [13], who modified an idea on math.stackexchange [1], and in a slightly less elegant form it appeared much earlier in [17]. For t R, set F (t) e t (1+x ) 1 + x dx. Then F () dx/(1 + x ) π/ and F ( ). Differentiating under the integral sign, F (t) Make the substitution y tx, with dy t dx, so te t (1+x ) dx te t e (tx) dx. F (t) e t e y dy Je t. For b >, integrate both sides from to b and use the Fundamental Theorem of Calculus: b F (t) dt J Letting b in the last equation, b e t dt F (b) F () J π J J π 4 J π. b e t dt. 5. Fifth Proof: A volume integral Our next proof is due to T. P. Jameson [4] and it was rediscovered by A. L. Delgado []. Revolve the curve z e 1 x in the xz-plane around the z-axis to produce the bell surface z e 1 (x +y ). See below, where the z-axis is vertical and passes through the top point, the x-axis lies just under the surface through the point in front, and the y-axis lies just under the surface through the point on the left. We will compute the volume V below the surface and above the xy-plane in two ways. First we compute V by horizontal slices, which are discs: V A(z) dz where A(z) is the area of the disc formed by slicing the surface at height z. Writing the radius of the disc at height z as r(z), A(z) πr(z). To compute r(z), the surface cuts the xz-plane at a pair of points (x, e 1 x ) where the height is z, so e 1 x z. Thus x ln z. Since x is the distance of these points from the z-axis, r(z) x ln z, so A(z) πr(z) πln z. Therefore V π ln z dz π (z ln z z) 1 π( 1 lim z ln z). z + By L Hospital s rule, lim z + z ln z, so V π. (A calculation of V by shells is in [1].) Next we compute the volume by vertical slices in planes x constant. Vertical slices are scaled bell curves: look at the black contour lines in the picture. The equation of the bell curve along the top of the vertical slice with x-coordinate x is z e 1 (x +y ), where y varies and x is fixed. Then

4 4 KEITH CONRAD V Thus V A(x) dx, where A(x) is the area of the x-slice: A(x) A(x) dx e 1 (x +y ) dy e 1 x e 1 y dy e 1 x I. e 1 x I dx I e 1 x dx I. Comparing the two formulas for V, we have π I, so I π. For any integer n, we have n! 6. Sixth Proof: The Γ-function Γ(x) t n e t dt. For x > we define t x t dt e t, so Γ(n) (n 1)! when n 1. Using integration by parts, Γ(x + 1) xγ(x). One of the basic properties of the Γ-function [14, pp ] is (6.1) Γ(x)Γ(y) Γ(x + y) t x 1 (1 t) y 1 dt.

5 Set x y 1/: Note Γ ( ) 1 te t dt t THE GAUSSIAN INTEGRAL 5 Γ ( ) 1 dt. t(1 t) e t dt t e x so 4J dt/ t(1 t). With the substitution t sin θ, 4J π/ sin θ cos θ dθ sin θ cos θ x x dx π π, e x dx J, so J π/. Equivalently, Γ(1/) π. Any method that proves Γ(1/) π is also a method that calculates e x dx. 7. Seventh Proof: Asymptotic estimates We will show J π/ by a technique whose steps are based on [15, p. 371]. For x, power series expansions show 1 + x e x 1/(1 x). Reciprocating and replacing x with x, we get (7.1) 1 x e x x. for all x R. For any positive integer n, raise the terms in (7.1) to the nth power and integrate from to 1: (1 x ) n dx e nx dx dx (1 + x ) n. Under the changes of variables x sin θ on the left, x y/ n in the middle, and x tan θ on the right, (7.) π/ (cos θ) n+1 dθ 1 n e y dy n Set I k π/ (cos θ) k dθ, so I π/, I 1 1, and (7.) implies (7.3) π/4 n nin+1 e y dy ni n. (cos θ) n dθ. We will show that as k, kik π/. Then n nin+1 n + 1In+1 1 π π n + 1 and so by (7.3) n nin e y dy π/. Thus J π/. n n n In 1 π π,

6 6 KEITH CONRAD To show ki k π/, first we compute several values of I k explicitly by a recursion. Using integration by parts, so I k π/ (cos θ) k dθ π/ (7.4) I k k 1 k I k. (cos θ) k 1 cos θ dθ (k 1)(I k I k ), Using (7.4) and the initial values I π/ and I 1 1, the first few values of I k are computed and listed in Table 1. k I k k I k π/ 1 1 (1/)(π/) 3 /3 4 (3/8)(π/) 5 8/15 6 (15/48)(π/) 7 48/15 Table 1. From Table 1 we see that (7.5) I n I n+1 1 π n + 1 for n 3, and this can be proved for all n by induction using (7.4). Since cos θ 1 for θ [, π/], we have I k I k 1 I k k k 1 I k by (7.4), so I k 1 I k as k. Therefore (7.5) implies as n. Then In 1 π n (n)i n π (n + 1)I n+1 (n)i n π as n, so ki k π/ as k. This completes our proof that J π/. Remark 7.1. This proof is closely related to the fifth proof using the Γ-function. Indeed, by (6.1) Γ( k+1 )Γ( 1 ) Γ( k ) t (k+1)/+1 (1 t) 1/ 1 dt,

7 THE GAUSSIAN INTEGRAL 7 and with the change of variables t (cos θ) for θ π/, the integral on the right is equal to π/ (cos θ) k dθ I k, so (7.5) is the same as π I n I n+1 (n + 1) Γ( n+1 )Γ( 1 ) Γ( n+ ) n+1 Γ( )Γ( 1 ) 4Γ( n+1 + 1) n+1 Γ( )Γ( 1 ) 4 n+1 Γ( n+1 ) Γ( 1 ) (n + 1), Γ( n+ )Γ( 1 ) Γ( n+3 ) or equivalently Γ(1/) π. We saw in the fifth proof that Γ(1/) π if and only if J π/. 8. Eighth Proof: The original proof The original proof that J π/ is due to Laplace [7] in (An English translation of Laplace s article is mentioned in the bibliographic citation for [7], with preliminary comments on that article in [16].) He wanted to compute (8.1) dx log x. Setting y log x, this integral is e y dy J, so we expect (8.1) to be π. Laplace s starting point for evaluating (8.1) was a formula of Euler: (8.) x r dx 1 x s x s+r dx 1 π 1 x s s(r + 1) for positive r and s. (Laplace himself said this formula held whatever be r or s, but if s < then the number under the square root is negative.) Accepting (8.), let r in it to get (8.3) dx 1 x s x s dx 1 π 1 x s s. Now let s in (8.3). Then 1 x s s log x by L Hopital s rule, so (8.3) becomes ( ) dx π. log x Thus (8.1) is π. Euler s formula (8.) looks mysterious, but we have met it before. In the formula let x s cos θ with θ π/. Then x (cos θ) 1/s, and after some calculations (8.) turns into (8.4) π/ (cos θ) (r+1)/s 1 dθ π/ (cos θ) (r+1)/s dθ 1 π (r + 1)/s. We used the integral I k π/ (cos θ) k dθ before when k is a nonnegative integer. This notation makes sense when k is any positive real number, and then (8.4) assumes the form I α I α+1 1 π α+1 for α (r +1)/s 1, which is (7.5) with a possibly nonintegral index. Letting r and s 1/(n+1)

8 8 KEITH CONRAD in (8.4) recovers (7.5). Letting s in (8.3) corresponds to letting n in (7.5), so the 6th proof is in essence a more detailed version of Laplace s 1774 argument. We will calculate 9. Ninth Proof: Contour Integration e x / dx using contour integrals and the residue theorem. However, we can t just integrate e z /, as this function has no poles. For a long time nobody knew how to handle this integral using contour integration. For instance, in 1914 Watson [18, p. 79] wrote Cauchy s theorem cannot be employed to evaluate all definite integrals; thus e x dx has not been evaluated except by other methods. In the 194s several contour integral solutions were published using awkward contours such as parallelograms [9], [11, Sect. 5] (see [1, Exer. 9, p. 113] for a recent appearance). Our approach will follow Kneser [5, p. 11] (see also [1, pp ] or []), using a rectangular contour and the function e z / 1 e π(1+i)z. This function comes out of nowhere, so our first task is to motivate the introduction of this function. We seek a meromorphic function f(z) to integrate around the rectangular contour γ R in the figure below, with vertices at R, R, R + ib, and R + ib, where b will be fixed and we let R. Suppose f(z) along the right and left sides of γ R uniformly as R. Then by applying the residue theorem and letting R, we would obtain (if the integrals converge) f(x) dx + f(x + ib) dx πi a Res za f(z), where the sum is over poles of f(z) with imaginary part between and b. This is equivalent to Therefore we want f(z) to satisfy (f(x) f(x + ib)) dx πi a (9.1) f(z) f(z + ib) e z /, Res za f(z). where f(z) and b need to be determined. Let s try f(z) e z / /d(z), for an unknown denominator d(z) whose zeros are poles of f(z). We want f(z) to satisfy (9.) f(z) f(z + τ) e z /

9 THE GAUSSIAN INTEGRAL 9 for some τ (which will not be purely imaginary, so (9.1) doesn t quite work, but (9.1) is only motivation). Substituting e z / /d(z) for f(z) in (9.) gives us ( ) / (9.3) e z / 1 e τz τ e z /. d(z) d(z + τ) Suppose d(z + τ) d(z). Then (9.3) implies d(z) 1 e τz τ /, and with this definition of d(z), f(z) satisfies (9.) if and only if e τ 1, or equivalently τ πiz. The simplest nonzero solution is τ π(1 + i). From now on this is the value of τ, so e τ / e iπ 1 and then f(z) e z / d(z) e z / 1 + e τz, which is Kneser s function mentioned earlier. This function satisfies (9.) and we henceforth ignore the motivation (9.1). Poles of f(z) are at odd integral multiples of τ/. We will integrate this f(z) around the rectangular contour γ R below, whose height is Im(τ). The poles of f(z) nearest the origin are plotted in the figure; they lie along the line y x. The only pole of f(z) inside γ R (for R > π/) is at τ/, so by the residue theorem e τ /8 f(z) dz πires zτ/ f(z) πi γ R ( τ)e τ / πie3τ /8 π(1 + i) π. As R, the value of f(z) tends to uniformly along the left and right sides of γ R, so π +i π f(x) dx + f(x) dx +i π f(z) dz f(x + i π) dx.

10 1 KEITH CONRAD In the second integral, write i π as τ π and use (real) translation invariance of dx to obtain π f(x) dx f(x + τ) dx Besides the integral formula (f(x) f(x + τ)) dx e x / dx by (9.). 1. Tenth Proof: Stirling s Formula in mathematics where π appears is in Stirling s formula: e 1 x dx π that we have been discussing, another place n! nn e n πn as n. In 173 De Moivre proved n! C(n n /e n ) n for some positive number C without being able to determine C. Stirling soon thereafter showed C π and wound up having the whole formula named after him. We will show that determining that the constant C in Stirling s formula is π is equivalent to showing that J π/ (or, equivalently, that I π). Applying (7.4) repeatedly, I n n 1 n I n (n 1)(n 3) I n 4 (n)(n ). (n 1)(n 3)(n 5) (5)(3)(1) I. (n)(n )(n 4) (6)(4)() Inserting (n )(n 4)(n 6) (6)(4)() in the top and bottom, I n (n 1)(n )(n 3)(n 4)(n 5) (6)(5)(4)(3)()(1) π (n)((n )(n 4) (6)(4)()) (n 1)! π n( n 1 (n 1)!). Applying De Moivre s asymptotic formula n! C(n/e) n n,, I n C((n 1)/e) n 1 n 1 π n( n 1 C((n 1)/e) n 1 n 1) (n 1) n 1 n 1 n 1 n (n 1) Ce(n 1) n 1 (n 1) (n 1) as n. For any a R, (1 + a/n) n e a as n, so (n + a) n e a n n. Substituting this into the above formula with a 1 and n replaced by n, (1.1) I n e 1 (n) n 1 n π n (n 1) Ce(e 1 n n ) 1 n π n C n. Since I k 1 I k, the outer terms in (7.3) are both asymptotic to ni n π/(c ) by (1.1). Therefore n e y dy π C π

11 THE GAUSSIAN INTEGRAL 11 as n, so J π/(c ). Therefore C π if and only if J π/. 11. Eleventh Proof: Fourier transforms For a continuous function f : R C that is rapidly decreasing at ±, its Fourier transform is the function Ff : R C defined by (Ff)(y) For example, (Ff)() f(x) dx. Here are three properties of the Fourier transform. f(x)e ixy dx. If f is differentiable, then after using differentiation under the integral sign on the Fourier transform of f we obtain (Ff) (y) ixf(x)e ixy dx i(f(xf(x)))(y). Using integration by parts on the Fourier transform of f, with u f(x) and dv e ixy dx, we obtain F(f )(y) iy(ff)(y). If we apply the Fourier transform twice then we recover the original function up to interior and exterior scaling: (11.1) (F f)(x) πf( x). Let s show the appearance of π in (11.1) is equivalent to the evaluation of I as π. Fixing a >, set f(x) e ax, so f (x) axf(x). Applying the Fourier transform to both sides of this equation implies iy(ff)(y) a 1 i (Ff) (y), which simplifies to (Ff) (y) 1 a y(ff)(y). The general solution of g (y) 1 ayg(y) is g(y) Ce y /(4a), so (Ff)(y) Ce y /(4a) for some constant C. Letting a 1, so f(x) e x /, we obtain (Ff)(y) Ce y / Cf(y). Setting y, the left side is (Ff)() e x / dx I, so I Cf() C. Applying the Fourier transform to both sides of the equation (Ff)(y) Cf(y), we get πf( x) C(Ff)(x) C f(x). At x this becomes π C, so I C ± π. Since I >, the number I is π. If we didn t know the constant on the right side of (11.1) were π, whatever its value is would wind up being C, so saying π appears on the right side of (11.1) is equivalent to saying I π.

12 1 KEITH CONRAD References [1] C. A. Berenstein and R. Gay, Complex Variables, Springer-Verlag, New York, [] A. L. Delgado, A Calculation of e x dx, The College Math. J. 34 (3), [3] H. Iwasawa, Gaussian Integral Puzzle, Math. Intelligencer 31 (9), [4] T. P. Jameson, The Probability Integral by Volume of Revolution, Mathematical Gazette 78 (1994), [5] H. Kneser, Funktionentheorie, Vandenhoeck and Ruprecht, [6] P. S. Laplace, Théorie Analytique des Probabilités, Courcier, 181. [7] P. S. Laplace, Mémoire sur la probabilité des causes par les évènemens, Oeuvres Complétes 8, (English trans. by S. Stigler as Memoir on the Probability of Causes of Events, Statistical Science 1 (1986), ) [8] P. M. Lee, [9] L. Mirsky, The Probability Integral, Math. Gazette 33 (1949), 79. Online at [1] C. P. Nicholas and R. C. Yates, The Probability Integral, Amer. Math. Monthly 57 (195), [11] G. Polya, Remarks on Computing the Probability Integral in One and Two Dimensions, pp in Berkeley Symp. on Math. Statist. and Prob., Univ. California Press, [1] R. Remmert, Theory of Complex Functions, Springer-Verlag, [13] M. Rozman, Evaluate Gaussian integral using differentiation under the integral sign, Course notes for Physics 4 (UConn), Spring 16. [14] W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, [15] M. Spivak, Calculus, W. A. Benjamin, [16] S. Stigler, Laplace s 1774 Memoir on Inverse Probability, Statistical Science 1 (1986), [17] J. van Yzeren, Moivre s and Fresnel s Integrals by Simple Integration, Amer. Math. Monthly 86 (1979), [18] G. N. Watson, Complex Integration and Cauchy s Theorem, Cambridge Univ. Press, Cambridge, [19] [] [1]