Differentiation under the integral sign

If $g(t,x)$ and its partial derivative $D_{i}g(t,x)$ are continuous for $t$ in a compact interval and $x$ in an open set, thenApply the fundamental theorem of calculus to the difference quotient in the $i$th coordinate and use uniform continuity on a compact neighbourhood.

Let $t_0$ be interior to a parameter interval. Suppose $h(t,\cdot )$ is a measurable function for each $t$, $h(t_0,\cdot )$ is a Lebesgue integrable function, and $h(\cdot ,x)$ is differentiable near $t_0$ for $x$ almost everywhere. If one integrable nonnegative function $D$ satisfies $|{\partial }_{t}h(t,x)|\le D(x)$ throughout that neighbourhood off one fixed null set, then $F(t)={\int }_{X}h(t,x)d\mu (x)$ is defined there and satisfies the displayed derivative formula at $t_0$.

The mean value theorem gives $|h(t,x)-h(t_0,x)|\le |t-t_0|D(x)$, proving local integrability. It also bounds the difference quotient by $D$. That quotient converges almost everywhere to ${\partial }_{t}h(t_0,x)$, so the dominated convergence theorem permits passing the limit through the Lebesgue integral.

Suppose $f$ and its partial derivative ${\partial }_{t}f$ are jointly continuous on a compact rectangle $J\times [a,b]$. Then $F(t)={\int }_a^{b}f(t,x)dx$ is continuously differentiable in the interior of $J$, and the displayed formula holds, with one-sided derivatives at parameter endpoints where the corresponding partial derivative is given.

To prove it, the extreme value theorem bounds $|{\partial }_{t}f|$ by $M_1$. For each fixed $x$, the mean value theorem bounds the difference quotient $[f(t+h,x)-f(t,x)]/h$ by $M_1$. It converges pointwise to ${\partial }_{t}f(t,x)$, and the constant $M_1$ is integrable over $[a,b]$. Dominated convergence gives the derivative formula. Apply continuity of parameter integrals on compact rectangles to ${\partial }_{t}f$ to obtain continuity of $F^{\prime }$. The proof works locally on unbounded parameter intervals as well.

For every real $t$, the mean value theorem bounds $|e^{tx}-1|\le |t|x{e}^{|t|}$ on $0<x<1$. Hence the apparent singularity at zero cancels, and the remaining bound $|t|e^{|t|}/\sqrt{1-x}$ is Lebesgue integrable. On a bounded parameter interval $|t|\le M$, every positive-order parameter derivative of the integrand is bounded in absolute value by $e^M/\sqrt{1-x}$. Repeated dominated differentiation under the integral sign therefore proves that $F$ is a smooth function, withIn particular, $F(0)=0$, $F^{\prime }(0)=2$, and $F^{\prime \prime }(t)>0$. Thus $F$ is strictly convex, lies above its tangent $2t$ at zero, and satisfies $F^{\prime }(t)>2$ for $t>0$ and $F^{\prime }(t)<2$ for $t<0$. The inequalities follow directly as well by comparing $e^{tx}$ to one. The substitution $x=1-s^2$ removes the square-root endpoint singularity for numerical integration:

For real $m>-1$ and integer $n\ge 0$, start from ${\int }_0^1{x}^{m}dx=1/(m+1)$ and differentiate $n$ times with respect to $m$. Each derivative adds one factor $\log x$, givingTo justify differentiation under the integral sign, restrict $m$ to a compact interval with lower endpoint $m_0>-1$. The derivatives are dominated by $x^{m_0}|\log x{|}^n$, which is integrable: $x=e^{-t}$ turns its integral into ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$. A power of $t$ is integrable against this decaying exponential. This also explains why $m=-1$ is the convergence threshold.

For every real $a$, define $I(a)={\int }_0^1\sin (a\log x)/\log xdx$. The apparent endpoint singularity at $x=1$ is removable, with limit $a$, and the bound $|\sin (a\log x)/\log x|\le |a|$ ensures integrability. Differentiation under the integral sign is justified by the integrable bound $|\cos (a\log x)|\le 1$ on the parameter derivative. Thuswhere $x=e^{-t}$ and the last integral is a Laplace transform. Since $I(0)=0$, integrating with respect to $a$ gives $I(a)=\arctan a$. This obtains an integral with a logarithmic denominator by differentiating away that denominator, then restoring the parameter dependence with one elementary antiderivative.

For real $\alpha$, the displayed improper integral converges and is an even function of $\alpha$. For $\alpha >0$, differentiation under the integral sign and partial fraction decomposition giveFor $\alpha \ne 1$, use $x^2/[(1+{\alpha }^2{x}^2)(1+x^2)]=[1/(1+x^2)-1/(1+{\alpha }^2{x}^2)]/({\alpha }^2-1)$. Integrate the resulting rational functions and extend to $\alpha =1$ by continuity. Integrating in $\alpha$ and using $E(0)=0$ gives $E(\alpha )=\pi \ln (1+\alpha )$ for $\alpha \ge 0$; evenness gives the formula for negative parameters. Thus $E^{\prime }(\alpha )=\pi \mathrm{sgn}(\alpha )/(1+|\alpha |)$ away from zero. At zero the one-sided derivatives are opposite, so $E$ is continuous but not differentiable. Substituting $\alpha =0$ into the pointwise differentiated integrand would therefore be an invalid interchange.

For every real parameter $a$,where $F(0)=0$. For $a>0$, differentiation under the integral sign and a partial fraction decomposition giveSolving this first-order linear differential equation yields $F(a)=a[\pi \ln (1+a)+C]$. To fix the constant, dominated convergence gives ${\lim }_{a\to 0^{+}}F(a)/a={\int }_0^{\infty }\ln (x^2)/(1+x^2)dx=0$; the last absolutely convergent integral changes sign under $x=1/t$. Thus $C=0$. Oddness in $a$ supplies the result for negative parameters. The initial value $F(0)=0$ alone does not determine $C$, since every solution $a[\pi \ln (1+a)+C]$ tends to zero as $a\to 0^{+}$.

For differentiable endpoints $a,b$ and a function $f$ with suitable continuous partial derivative in the parameter, the moving-endpoint rule isTo prove it locally, put $H(x,u)={\int }_c^{u}f(x,t)dt$ for a fixed reference endpoint $c$. The original integral is $H(x,b(x))-H(x,a(x))$. Apply the chain rule, use the fundamental theorem of calculus for $H_u=f$, and use differentiation under the integral sign for $H_x$. The two boundary terms account for the moving endpoints, while the remaining integral accounts for variation of the integrand with the parameter. For improper integrals, convergence and a suitable domination condition must be verified before interchanging the derivative and integral.

On a star-shaped neighbourhood of $a$, a smooth function satisfiesThe fundamental theorem of calculus along the segment proves the identity, and differentiation under the integral sign proves smoothness of the coefficients.

For a smooth function $f(x,y)$,where the smooth remainder is