Differentiation under the integral sign

Some definite integrals refuse to yield an elementary antiderivative — you can write down the integral, but no combination of familiar functions equals it. Differentiation under the integral sign, also called Leibnitz's rule for integrals, is the escape route. Instead of attacking the integral directly, you let one of its constants become a variable parameter $\alpha$ , differentiate the resulting integral with respect to $\alpha$ , and hope the differentiated integrand is easier than the original. Often it collapses to something elementary. You then integrate that simpler result back with respect to $\alpha$ and pin the constant of integration using one convenient value where the original integral is trivially known. This article states the general rule (which also handles parameter-dependent limits of integration), its important fixed-limit special case, and walks through the technique on the same integral used throughout this unit.

The core insight

Promoting a constant in a hard integral to a parameter $\alpha$ , differentiating under the integral sign, and integrating the simplified result back turns an integral with no elementary antiderivative into one that can be evaluated exactly.

The core idea

Setting up the parameter-dependent integral. Let $f(x,\alpha)$ depend on $x$ and on a parameter $\alpha$ , and let the limits of integration $a(\alpha)$ and $b(\alpha)$ possibly depend on $\alpha$ too. Define

I(\alpha) = \int_{a(\alpha)}^{b(\alpha)} f(x,\alpha)\,dx.

where: $f(x,\alpha)$ is the integrand, $x$ is the variable being integrated out, and $\alpha$ is the parameter surviving in $I(\alpha)$ .

Leibnitz's rule (general form). Differentiating $I(\alpha)$ with respect to $\alpha$ gives

\frac{d}{d\alpha}\int_{a(\alpha)}^{b(\alpha)} f(x,\alpha)\,dx = \int_{a(\alpha)}^{b(\alpha)} \frac{\partial f}{\partial\alpha}\,dx \;+\; f\big(b(\alpha),\alpha\big)\,b'(\alpha) \;-\; f\big(a(\alpha),\alpha\big)\,a'(\alpha).

where: the first term is the change coming from the integrand itself shifting with $\alpha$ , and the remaining two terms are the change coming from the upper and lower limits sweeping in or out as $\alpha$ varies — the upper limit contributes with a plus sign, the lower limit with a minus sign.

Fixed-limit special case. When $a$ and $b$ are constants, both boundary terms vanish and

\frac{d}{d\alpha}\int_a^b f(x,\alpha)\,dx = \int_a^b \frac{\partial f}{\partial \alpha}\,dx,

i.e. differentiation and integration can simply be swapped — "differentiate inside the integral sign." This is the version used most often as an evaluation technique: introduce a parameter into a hard integral, differentiate with respect to it, integrate the simpler result back, and fix the constant using a value of $\alpha$ where the original integral is trivial (frequently $\alpha = 0$ ). The figure below shows the geometric picture behind the general rule: as $\alpha$ changes, the region under $f(x,\alpha)$ changes both because the curve itself moves and because its left and right edges slide.

Figure

Two shaded regions under a curve for two nearby values of alpha showing the curve shifting and both boundaries sliding outward

Leibnitz rule from the multivariable chain rule

The general rule is a direct consequence of the fundamental theorem of calculus applied through a multivariable chain rule, since $I(\alpha)$ really depends on $\alpha$ in three separate places at once: through the lower limit, the upper limit, and the integrand — and disentangling those three simultaneous dependences using the ordinary chain rule for functions of several variables is exactly what proves the formula stated above.

Derivation ALeibnitz's rule from the chain rule

The goal is to prove $\dfrac{d}{d\alpha}\displaystyle\int_{a(\alpha)}^{b(\alpha)} f(x,\alpha)\,dx = \displaystyle\int_{a(\alpha)}^{b(\alpha)} \dfrac{\partial f}{\partial\alpha}\,dx + f(b(\alpha),\alpha)\,b'(\alpha) - f(a(\alpha),\alpha)\,a'(\alpha)$ by treating $I(\alpha)$ as a function built from three independent slots — a lower limit, an upper limit, and the parameter sitting inside the integrand — and applying the ordinary multivariable chain rule to it.

Step 1. Introduce a three-slot helper function. Define $F(u,v,\alpha) = \displaystyle\int_u^v f(x,\alpha)\,dx$ , treating $u$ , $v$ and $\alpha$ as three independent variables. Then the original object is the composite $I(\alpha) = F\big(a(\alpha),\,b(\alpha),\,\alpha\big)$ , where $u=a(\alpha)$ and $v=b(\alpha)$ are themselves functions of $\alpha$ .

Step 2. Find the partial derivatives of $F$ with respect to $u$ and $v$ using the fundamental theorem of calculus. Holding $\alpha$ fixed, $\dfrac{\partial F}{\partial v} = f(v,\alpha)$ directly from the fundamental theorem, and $\dfrac{\partial F}{\partial u} = -f(u,\alpha)$ because reversing the direction of integration flips the sign: $\displaystyle\int_u^v f\,dx = -\int_v^u f\,dx$ .

Step 3. Find the partial derivative of $F$ with respect to its third slot, holding $u$ and $v$ fixed. Since the limits are frozen, only the integrand changes with $\alpha$ , so $\dfrac{\partial F}{\partial \alpha} = \displaystyle\int_u^v \dfrac{\partial f}{\partial \alpha}\,dx$ — this is exactly the fixed-limit special case, applied here as one piece of the larger proof, and it is valid whenever $f$ and $\partial f/\partial\alpha$ are continuous on the region of integration.

Step 4. Apply the multivariable chain rule to the composite $I(\alpha) = F(a(\alpha),b(\alpha),\alpha)$ : differentiating a function of three slots, each of which may depend on $\alpha$ , gives $\dfrac{dI}{d\alpha} = \dfrac{\partial F}{\partial u}\,a'(\alpha) + \dfrac{\partial F}{\partial v}\,b'(\alpha) + \dfrac{\partial F}{\partial \alpha}$ . Substituting the three partial derivatives from Steps 2 and 3 gives $\dfrac{dI}{d\alpha} = -f(a(\alpha),\alpha)\,a'(\alpha) + f(b(\alpha),\alpha)\,b'(\alpha) + \displaystyle\int_{a(\alpha)}^{b(\alpha)} \dfrac{\partial f}{\partial \alpha}\,dx$ , which is Leibnitz's rule exactly as stated, with the lower-limit term carrying the minus sign inherited from Step 2.

Worked example

Problem. Evaluate $I(\alpha) = \displaystyle\int_0^1 \frac{x^\alpha - 1}{\ln x}\,dx$ for $\alpha > -1$ , using differentiation under the integral sign. This integral has no elementary closed form as it stands, so it cannot be attacked by ordinary antidifferentiation in $x$ .

Step 1. Differentiate with respect to $\alpha$ under the fixed-limit rule (the limits $0$ and $1$ are constants, so only the integrand changes): $I'(\alpha) = \displaystyle\int_0^1 \dfrac{\partial}{\partial \alpha}\left(\dfrac{x^\alpha - 1}{\ln x}\right)dx$ .

Step 2. Compute the partial derivative inside the integral. Since $\dfrac{\partial}{\partial \alpha}\left(x^\alpha\right) = x^\alpha \ln x$ (treating $x$ as constant), the $\ln x$ in the numerator cancels the $\ln x$ in the denominator: $I'(\alpha) = \displaystyle\int_0^1 \dfrac{x^\alpha \ln x}{\ln x}\,dx = \displaystyle\int_0^1 x^\alpha\,dx = \dfrac{1}{\alpha+1}$ . The parameter-dependence has become completely elementary.

Step 3. Integrate $I'(\alpha)$ back with respect to $\alpha$ : $I(\alpha) = \ln(\alpha+1) + C$ , for some constant $C$ still to be found.

Step 4. Pin down $C$ using a convenient value of $\alpha$ where the original integral is directly computable. At $\alpha = 0$ , $I(0) = \displaystyle\int_0^1 \dfrac{x^0 - 1}{\ln x}\,dx = \displaystyle\int_0^1 0\,dx = 0$ . Substituting into Step 3's formula, $0 = \ln(1) + C = 0 + C$ , so $C = 0$ . Hence $\boxed{I(\alpha) = \ln(1+\alpha)}$ .

Competitive stretch

Once $I(\alpha) = \ln(1+\alpha)$ is known, particular numerical integrals fall out for free without repeating the method: setting $\alpha = 1$ gives $\displaystyle\int_0^1 \dfrac{x-1}{\ln x}\,dx = \ln 2$ , and setting $\alpha = 2$ gives $\displaystyle\int_0^1 \dfrac{x^2-1}{\ln x}\,dx = \ln 3$ . This is the real payoff of the technique — one differentiation-and-integration cycle solves an entire family of integrals indexed by $\alpha$ , not just the one you started with.

Common traps

Using the fixed-limit shortcut (differentiate only the integrand) when the limits actually depend on $\alpha$ — in that case the full Leibnitz rule with both boundary terms is required.
Getting the sign wrong on the lower-limit boundary term: it always enters with a minus sign, $-f(a(\alpha),\alpha)\,a'(\alpha)$ , while the upper-limit term is plus.
Confusing partial and total derivatives — differentiating the integrand means $\partial f/\partial\alpha$ with $x$ held constant, never a total derivative in $x$ .
After finding $I'(\alpha)$ , forgetting the final step: integrate back with respect to $\alpha$ and use a specific, directly computable value of $\alpha$ (often $\alpha=0$ ) to pin down the constant of integration — skipping this leaves $I(\alpha)$ undetermined up to a constant.

Ask the tutor: if choosing which constant to promote to a parameter, or keeping the boundary-term signs straight in the full Leibnitz rule, feels shaky, ask the AI tutor to walk through one or two more parameter-introduction problems with you.

Why does differentiating a hard integral ever make it easier instead of harder?

Differentiating with respect to the parameter $\alpha$ only touches the piece of the integrand that actually depends on $\alpha$ , and quite often that piece simplifies dramatically — in the worked example, $\partial(x^\alpha)/\partial\alpha = x^\alpha\ln x$ exactly cancelled the awkward $\ln x$ in the denominator. The integral in $x$ that remains is frequently elementary even though the original integral, still containing $\ln x$ in an unremovable way, was not.

How do I decide whether to use the fixed-limit shortcut or the full Leibnitz rule?

Check whether the limits of integration depend on the variable you are differentiating with respect to. If both limits are plain numbers or constants unrelated to $\alpha$ , only the integrand changes as $\alpha$ varies, so the fixed-limit version applies and you differentiate straight through the integral sign. If either limit is itself a function of $\alpha$ , you must include its boundary term, with the correct plus or minus sign, or the answer will be missing a piece.

When I am given a hard integral with no parameter at all, how do I choose one to introduce?

Look for a constant in the integrand that, if replaced by a variable $\alpha$ , would make differentiation cancel or simplify an awkward factor — commonly an exponent, as in $x^\alpha$ , or a coefficient inside a trigonometric or exponential term. A good choice is one where $\partial f/\partial \alpha$ removes the exact feature (often a logarithm or a factor of $x$ ) that was blocking direct integration. It also helps to pick the parameter so that some specific value, frequently $\alpha=0$ , makes the original integral trivial to evaluate directly, since that value is what fixes your constant of integration at the end.