Keith Conrad

1. Introduction

The method of differentiation under the integral sign, due to Leibniz in 1697 [4], concerns integrals depending on a parameter, such as $\int_{0}^{1} x^2 e^{-t x} \mathrm{~d} x$. Here $t$ is the extra parameter. (Since $x$ is the variable of integration, $x$ is not a parameter.) In general, we might write such an integral as \[\tag{1.1} \int_{a}^{b} f(x, t) \mathrm{d} x, \] where $f(x, t)$ is a function of two variables like $f(x, t)=x^2 e^{-t x}$.
Example 1.1. Let $f(x, t)=\left(2 x+t^{3}\right)^2$. Then $\int_{0}^{1} f(x, t) \mathrm{d} x=\int_{0}^{1}\left(2 x+t^{3}\right)^2 \mathrm{~d} x$. An anti-derivative of $\left(2 x+t^{3}\right)^2$ with respect to $x$ is $\frac{1}{6}\left(2 x+t^{3}\right)^{3}$, so \[ \int_{0}^{1}\left(2 x+t^{3}\right)^2 \mathrm{~d} x=\left.\frac{\left(2 x+t^{3}\right)^{3}}{6}\right|_{x=0} ^{x=1}=\frac{\left(2+t^{3}\right)^{3}-t^{9}}{6}=\frac{4}{3}+2 t^{3}+t^{6} . \] This answer is a function of $t$, which makes sense since the integrand depends on $t$. We integrate over $x$ and are left with something that depends only on $t$, not $x$.

An integral like $\int_{a}^{b} f(x, t) \mathrm{d} x$ is a function of $t$, so we can ask about its $t$-derivative, assuming that $f(x, t)$ is nicely behaved. The rule, called differentiation under the integral sign, is that the $t$-derivative of the integral of $f(x, t)$ is the integral of the $t$-derivative of $f(x, t)$ : \[\tag{1.2} \frac{\mathrm{d}}{\mathrm{d} t} \int_{a}^{b} f(x, t) \mathrm{d} x=\int_{a}^{b} \frac{\partial}{\partial t} f(x, t) \mathrm{d} x . \] If you are used to thinking mostly about functions with one variable, not two, keep in mind that (1.2) involves integrals and derivatives with respect to separate variables: integration with respect to $x$ and differentiation with respect to $t$.

Example 1.2. We saw in Example $1.1$ that $\int_{0}^{1}\left(2 x+t^{3}\right)^2 \mathrm{~d} x=4 / 3+2 t^{3}+t^{6}$, whose $t$-derivative is $6 t^2+6 t^{5}$. According to (1.2), we can also compute the $t$-derivative of the integral like this: \[ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d} t} \int_{0}^{1}\left(2 x+t^{3}\right)^2 \mathrm{~d} x &=\int_{0}^{1} \frac{\partial}{\partial t}\left(2 x+t^{3}\right)^2 \mathrm{~d} x \\ &=\int_{0}^{1} 2\left(2 x+t^{3}\right)\left(3 t^2\right) \mathrm{d} x \\ &=\int_{0}^{1}\left(12 t^2 x+6 t^{5}\right) \mathrm{d} x \\ &=6 t^2 x^2+\left.6 t^{5} x\right|_{x=0} ^{x=1} \\ &=6 t^2+6 t^{5} \end{aligned} \] The answer agrees with our first, more direct, calculation. We will apply (1.2) to many examples of integrals, in Section 12 we will discuss the justification of this method in our examples, and then we'll give some more examples.

2. Euler’s factorial integral in a new light

For integers $n \geq 0$, Euler's integral formula for $n!$ is \[\tag{2.1} \int_{0}^{\infty} x^{n} e^{-x} \mathrm{~d} x=n !, \] which can be obtained by repeated integration by parts starting from the formula \[\tag{2.2} \int_{0}^{\infty} e^{-x} \mathrm{~d} x=1 \] when $n=0$. Now we are going to derive Euler's formula in another way, by repeated differentiation after introducing a parameter $t$ into (2.2).
For $t>0$, let $x=t u$. Then $\mathrm{d} x=t \mathrm{~d} u$ and (2.2) becomes \[ \int_{0}^{\infty} t e^{-t u} \mathrm{~d} u=1 . \] Dividing by $t$ and writing $u$ as $x$ (why is this not a problem?), we get \[\tag{2.3} \int_{0}^{\infty} e^{-t x} \mathrm{~d} x=\frac{1}{t} . \] This is a parametric form of (2.2), where both sides are now functions of $t$. We need $t>0$ in order that $e^{-t x}$ is integrable over the region $x \geq 0$.

Now we bring in differentiation under the integral sign. Differentiate both sides of (2.3) with respect to $t$, using (1.2) to treat the left side. We obtain \[ \int_{0}^{\infty}-x e^{-t x} \mathrm{~d} x=-\frac{1}{t^2} \] so $$\tag{2.4}\int_{0}^{\infty} x e^{-t x} \mathrm{~d} x=\frac{1}{t^2}$$ Differentiate both sides of $(2.4)$ with respect to $t$, again using (1.2) to handle the left side. We get \[ \int_{0}^{\infty}-x^2 e^{-t x} \mathrm{~d} x=-\frac{2}{t^{3}} . \] Taking out the sign on both sides, \[ \int_{0}^{\infty} x^2 e^{-t x} \mathrm{~d} x=\frac{2}{t^{3}} . \] If we continue to differentiate each new equation with respect to $t$ a few more times, we obtain \begin{aligned} &\int_{0}^{\infty} x^{3} e^{-t x} \mathrm{~d} x=\frac{6}{t^{4}}, \\ &\int_{0}^{\infty} x^{4} e^{-t x} \mathrm{~d} x=\frac{24}{t^{5}}, \end{aligned} and \[ \int_{0}^{\infty} x^{5} e^{-t x} \mathrm{~d} x=\frac{120}{t^{6}} \] Do you see the pattern? It is \[ \int_{0}^{\infty} x^{n} e^{-t x} \mathrm{~d} x=\frac{n !}{t^{n+1}} . \] We have used the presence of the extra variable $t$ to get these equations by repeatedly applying $\mathrm{d} / \mathrm{d} t$. Now specialize $t$ to 1 in (2.6). We obtain \[ \int_{0}^{\infty} x^{n} e^{-x} \mathrm{~d} x=n ! \] which is our old friend (2.1). Voilà!
The idea that made this work is introducing a parameter $t$, using calculus on $t$, and then setting $t$ to a particular value so it disappears from the final formula. In other words, sometimes to solve a problem it is useful to solve a more general problem. Compare (2.1) to (2.6).

3. A Damped Sine Integral

We are going to use differentiation under the integral sign to prove \[ \int_{0}^{\infty} e^{-t x} \frac{\sin x}{x}\mathrm{~d} x=\frac{\pi}{2}-\arctan t \] for $t>0$. Call this integral $F(t)$ and set $f(x, t)=e^{-t x}(\sin x) / x$, so $(\partial / \partial t) f(x, t)=-e^{-t x} \sin x$. Then \[ F'(t)=-\int_{0}^{\infty} e^{-t x}(\sin x) \mathrm{~d} x \] The integrand $e^{-t x} \sin x$, as a function of $x$, can be integrated by parts: \[ \int e^{a x} \sin x \mathrm{~d} x=\frac{(a \sin x-\cos x)}{1+a^2} e^{a x} \] Applying this with $a=-t$ and turning the indefinite integral into a definite integral, \[ F'(t)=-\int_{0}^{\infty} e^{-t x}(\sin x) \mathrm{~d} x=\left.\frac{(t \sin x+\cos x)}{1+t^2} e^{-t x}\right|_{x=0} ^{x=\infty} \] As $x \rightarrow \infty, t \sin x+\cos x$ oscillates a lot, but in a bounded way (since $\sin x$ and $\cos x$ are bounded functions), while the term $e^{-t x}$ decays exponentially to 0 since $t>0$. So the value at $x=\infty$ is 0 . Therefore \[ F'(t)=-\int_{0}^{\infty} e^{-t x}(\sin x) \mathrm{~d} x=-\frac{1}{1+t^2} . \] We know an explicit antiderivative of $1 /\left(1+t^2\right)$, namely $\arctan t$. Since $F(t)$ has the same $t$-derivative as $-\arctan t$, they differ by a constant: for some number $C$, \[\tag{3.1} \int_{0}^{\infty} e^{-t x} \frac{\sin x}{x} \mathrm{~d} x=-\arctan t+C\quad \text { for } t>0 \] We've computed the integral, up to an additive constant, without finding an antiderivative of $e^{-t x}(\sin x) / x$

To compute $C$ in (3.1), let $t \rightarrow \infty$ on both sides. Since $|(\sin x) / x| \leq 1$, the absolute value of the integral on the left is bounded from above by $\int_{0}^{\infty} e^{-t x} \mathrm{~d} x=1 / t$, so the integral on the left in (3.1) tends to 0 as $t \rightarrow \infty$. Since arctan $t \rightarrow \pi / 2$ as $t \rightarrow \infty$, equation (3.1) as $t \rightarrow \infty$ becomes $0=-\frac{\pi}{2}+C$, so $C=\pi / 2$. Feeding this back into (3.1) \[\tag{3.2} \int_{0}^{\infty} e^{-t x} \frac{\sin x}{x} \mathrm{~d} x=\frac{\pi}{2}-\arctan t \quad \text { for } t>0 \] If we let $t \rightarrow 0^+$ in (3.2), this equation suggests that \[\tag{3.3} \int_{0}^{\infty} \frac{\sin x}{x} \mathrm{~d} x=\frac{\pi}{2} \] which is true and it is important in signal processing and Fourier analysis. It is a delicate matter to derive (3.3) from (3.2) since the integral in (3.3) is not absolutely convergent. Details are provided in an appendix.

4. The Gaussian integral

The improper integral formula \[\tag{4.1} \int_{-\infty}^{\infty} e^{-x^2 / 2} \mathrm{~d} x=\sqrt{2 \pi} \] is fundamental to probability theory and Fourier analysis. The function $\frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}$ is called a Gaussian, and (4.1) says the integral of the Gaussian over the whole real line is 1.

The physicist Lord Kelvin (after whom the Kelvin temperature scale is named) once wrote (4.1) on the board in a class and said "A mathematician is one to whom that [pointing at the formula] is as obvious as twice two makes four is to you." We will prove (4.1) using differentiation under the integral sign. The method will not make (4.1) as obvious as $2 \cdot 2=4$. If you take further courses you may learn more natural derivations of (4.1) so that the result really does become obvious. For now, just try to follow the argument here step-by-step.
We are going to aim not at (4.1), but at an equivalent formula over the range $x \geq 0$ : \[\tag{4.2} \int_{0}^{\infty} e^{-x^2 / 2} \mathrm{~d} x=\frac{\sqrt{2 \pi}}{2}=\sqrt{\frac{\pi}{2}} . \] Call the integral on the left $I$. For $t \in \mathbf{R}$, set \[ F(t)=\int_{0}^{\infty} \frac{e^{-t^2\left(1+x^2\right) / 2}}{1+x^2} \mathrm{~d} x . \] Then $F(0)=\int_{0}^{\infty}\mathrm{d}x /\left(1+x^2\right)=\pi / 2$ and $F(\infty)=0$. Differentiating under the integral sign, \[ F'(t)=\int_{0}^{\infty}-t e^{-t^2\left(1+x^2\right) / 2} \mathrm{~d} x=-t e^{-t^2 / 2} \int_{0}^{\infty} e^{-(t x)^2 / 2} \mathrm{~d} x . \] Make the substitution $y=t x$, with $\mathrm{d} y=t \mathrm{~d} x$, so \[ F'(t)=-e^{-t^2 / 2} \int_{0}^{\infty} e^{-y^2 / 2} \mathrm{~d} y=-I e^{-t^2 / 2} . \] For $b>0$, integrate both sides from 0 to $b$ and use the Fundamental Theorem of Calculus: \[ \int_{0}^{b} F'(t) \mathrm{~d} t=-I \int_{0}^{b} e^{-t^2 / 2} \mathrm{~d} t \Longrightarrow F(b)-F(0)=-I \int_{0}^{b} e^{-t^2 / 2} \mathrm{~d} t \] Letting $b \rightarrow \infty$, \[ 0-\frac{\pi}{2}=-I^2 \Longrightarrow I^2=\frac{\pi}{2} \Longrightarrow I=\sqrt{\frac{\pi}{2}} . \] I learned this from Michael Rozman [12], who modified an idea on a Math Stackexchange question [3], and in a slightly less elegant form it appeared much earlier in [15].

5. Higher moments of the Gaussian

For every integer $n \geq 0$ we want to compute a formula for \[\tag{5.1} \int_{-\infty}^{\infty} x^{n} e^{-x^2 / 2} \mathrm{~d} x \] (Integrals of the type $\int x^{n} f(x) \mathrm{d} x$ for $n=0,1,2, \ldots$ are called the moments of $f(x)$, so (5.1) is the $n$-th moment of the Gaussian.) When $n$ is odd, (5.1) vanishes since $x^{n} e^{-x^2 / 2}$ is an odd function. What if $n=0,2,4, \ldots$ is even? The first case, $n=0$, is the Gaussian integral (4.1): \[\tag{5.2} \int_{-\infty}^{\infty} e^{-x^2 / 2} \mathrm{~d} x=\sqrt{2 \pi} . \] To get formulas for (5.1) when $n \neq 0$, we follow the same strategy as our treatment of the factorial integral in Section 2: stick a $t$ into the exponent of $e^{-x^2/ 2}$ and then differentiate repeatedly with respect to $t$. For $t>0$, replacing $x$ with $\sqrt{t} x$ in (5.2) gives \[\tag{5.3} \int_{-\infty}^{\infty} e^{-t x^2 / 2} \mathrm{~d} x=\frac{\sqrt{2 \pi}}{\sqrt{t}} . \] Differentiate both sides of (5.3) with respect to $t$, using differentiation under the integral sign on the left: \[ \int_{-\infty}^{\infty}-\frac{x^2}{2} e^{-t x^2 / 2} \mathrm{~d} x=-\frac{\sqrt{2 \pi}}{2 t^{3 / 2}}, \] so \[\tag{5.4} \int_{-\infty}^{\infty} x^2 e^{-t x^2 / 2} \mathrm{~d} x=\frac{\sqrt{2 \pi}}{t^{3 / 2}} \] Differentiate both sides of (5.4) with respect to $t$. After removing a common factor of $-1 / 2$ on both sides, we get \[\tag{5.5} \int_{-\infty}^{\infty} x^{4} e^{-t x^2 / 2} \mathrm{~d} x=\frac{3 \sqrt{2 \pi}}{t^{5 / 2}} \] Differentiating both sides of (5.5) with respect to $t$ a few more times, we get \[ \begin{gathered} \int_{-\infty}^{\infty} x^{6} e^{-t x^2 / 2} \mathrm{~d} x=\frac{3 \cdot 5 \sqrt{2 \pi}}{t^{7 / 2}} \\ \int_{-\infty}^{\infty} x^{8} e^{-t x^2 / 2} \mathrm{~d} x=\frac{3 \cdot 5 \cdot 7 \sqrt{2 \pi}}{t^{9 / 2}} \end{gathered} \] and \[ \int_{-\infty}^{\infty} x^{10} e^{-t x^2 / 2} \mathrm{~d} x=\frac{3 \cdot 5 \cdot 7 \cdot 9 \sqrt{2 \pi}}{t^{11 / 2}} . \] Quite generally, when $n$ is even \[ \int_{-\infty}^{\infty} x^{n} e^{-t x^2 / 2} \mathrm{~d} x=\frac{1 \cdot 3 \cdot 5 \cdots(n-1)}{t^{(n+1) / 2}} \sqrt{2 \pi} \] where the numerator is the product of the positive odd integers from 1 to $n-1$ (understood to be the empty product 1 when $n=0)$. In particular, taking $t=1$ we have computed (5.1): \[ \int_{-\infty}^{\infty} x^{n} e^{-x^2 / 2} \mathrm{~d} x=1 \cdot 3 \cdot 5 \cdots(n-1) \sqrt{2 \pi} \] As an application of (5.4), we now compute $\left(\frac{1}{2}\right) !:=\int_{0}^{\infty} x^{1 / 2} e^{-x} \mathrm{~d} x$, where the notation $\left(\frac{1}{2}\right) !$ and its definition are inspired by Euler's integral formula (2.1) for $n!$ when $n$ is a nonnegative integer. Using the substitution $u=x^{1 / 2}$ in $\int_{0}^{\infty} x^{1 / 2} e^{-x} \mathrm{~d} x$, we have \[ \begin{aligned} \left(\frac{1}{2}\right) ! &=\int_{0}^{\infty} x^{1 / 2} e^{-x} \mathrm{~d} x \\ &=\int_{0}^{\infty} u e^{-u^2}(2 u) \mathrm{~d} u \\ &=2 \int_{0}^{\infty} u^2 e^{-u^2} \mathrm{~d} u \\ &=\int_{-\infty}^{\infty} u^2 e^{-u^2} \mathrm{~d} u \\ &=\frac{\sqrt{2 \pi}}{2^{3 / 2}} \text { by }(5.4) \text { at } t=2 \\ &=\frac{\sqrt{\pi}}{2} \end{aligned} \]

6. A Cosine Transform of the Gaussian

We are going to compute \[ F(t)=\int_{0}^{\infty} \cos (t x) e^{-x^2/2} \mathrm{~d} x \] by looking at its $t$-derivative: \[\tag{6.1} F'(t)=\int_{0}^{\infty}-x \sin (t x) e^{-x^2 / 2} \mathrm{~d} x . \] This is good from the viewpoint of integration by parts since $-x e^{-x^2 / 2}$ is the derivative of $e^{-x^2 / 2}$. So we apply integration by parts to (6.1): \[ u=\sin (t x), \quad \mathrm{d} v=-x e^{-x^2} \mathrm{~d} x \] and \[ \mathrm{d} u=t \cos (t x) \mathrm{d} x, \quad v=e^{-x^2 / 2} \] Then \begin{aligned} F'(t) &=\int_{0}^{\infty} u \mathrm{~d} v \\ &=\left.u v\right|_{0} ^{\infty}-\int_{0}^{\infty} v \mathrm{~d} u \\ &=\left.\frac{\sin (t x)}{e^{x^2 / 2}}\right|_{x=0} ^{x=\infty}-t \int_{0}^{\infty} \cos (t x) e^{-x^2 / 2} \mathrm{~d} x \\ &=\left.\frac{\sin (t x)}{e^{x^2 / 2}}\right|_{x=0} ^{x=\infty}-t F(t) \end{aligned} As $x \rightarrow \infty, e^{x^2 / 2}$ blows up while $\sin (t x)$ stays bounded, so $\sin (t x) / e^{x^2 / 2}$ goes to 0 . Therefore $F'(t)=-t F(t)$ We know the solutions to this differential equation: constant multiples of $e^{-t^2 / 2}$. So \[ \int_{0}^{\infty} \cos (t x) e^{-x^2 / 2} \mathrm{~d} x=C e^{-t^2 / 2} \] for some constant $C$. To find $C$, set $t=0$. The left side is $\int_{0}^{\infty} e^{-x^2 / 2} \mathrm{~d} x$, which is $\sqrt{\pi / 2}$ by (4.2). The right side is $C$. Thus $C=\sqrt{\pi / 2}$, so we are done: for all real $t$, \[ \int_{0}^{\infty} \cos (t x) e^{-x^2 / 2} \mathrm{~d} x=\sqrt{\frac{\pi}{2}} e^{-t^2 / 2} . \] Remark 6.1. If we want to compute $G(t)=\int_{0}^{\infty} \sin (t x) e^{-x^2 / 2} \mathrm{~d} x$, with $\sin (t x)$ in place of $\cos (t x)$, then in place of $F'(t)=-t F(t)$ we have $G'(t)=1-t G(t)$, and $G(0)=0$. From the differential equation, $\left(e^{t^2 / 2} G(t)\right)'=e^{t^2 / 2}$, so $G(t)=e^{-t^2 / 2} \int_{0}^{t} e^{x^2 / 2} \mathrm{~d} x$. So while $\int_{0}^{\infty} \cos (t x) e^{-x^2 / 2} \mathrm{~d} x=\sqrt{\frac{\pi}{2}} e^{-t^2 / 2}$, the integral $\int_{0}^{\infty} \sin (t x) e^{-x^2 / 2} \mathrm{~d} x$ is impossible to express in terms of elementary functions.

7. The Gaussian times a logarithm

We will compute \[ \int_{0}^{\infty}(\log x) e^{-x^2} \mathrm{~d} x . \] Integrability at $\infty$ follows from rapid decay of $e^{-x^2}$ at $\infty$, and integrability near $x=0$ follows from the integrand there being nearly $\log x$, which is integrable on $[0,1]$, so the integral makes sense. (This example was brought to my attention by Harald Helfgott.)