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Integration by differentiation under the integral sign

Method -> "DiffUnderInt" evaluates a parameter-dependent definite integral

\[ I(p) \;=\; \int_a^b f(x, p)\, dx \]

by the Leibniz integral rule — differentiating the integrand with respect to a parameter \(p\), integrating the (simpler) result, and integrating back. It is the technique Leibniz stated in the 1690s, that Euler wielded to evaluate hard integrals, and that Richard Feynman made famous in Surely You're Joking, Mr. Feynman! — where he calls it "a different box of tools" and puts it to work at Los Alamos. Its informal name, "Feynman's trick," dates from that book.

In[1]:= Integrate[(x^a - 1)/Log[x], {x, 0, 1}, Method -> "DiffUnderInt"]
Out[1]= Log[1 + a]

The integrand \((x^a-1)/\log x\) has no elementary antiderivative in \(x\), yet the answer is a clean logarithm. The trick: differentiate under the integral sign with respect to \(a\), and the integral collapses.

1. The mathematical idea

The Leibniz integral rule

For an integral with a parameter \(p\) and fixed limits, differentiation passes through the integral sign,

\[ \frac{d}{dp}\int_a^b f(x,p)\,dx \;=\; \int_a^b \frac{\partial f}{\partial p}(x,p)\,dx, \]

provided \(f\) and \(\partial_p f\) are continuous and the differentiated integral converges uniformly in \(p\) (a dominated-convergence condition). The power of the identity is that \(\partial_p f\) is very often simpler than \(f\) — an exponential loses its polynomial prefactor, a logarithm or arctangent turns into a rational function — so the integral on the right is one you can do.

The method as a first-order ODE

Write \(J(p)=\int_a^b \partial_p f\,dx\) for that simpler integral. Then \(I(p)\) satisfies the elementary ordinary differential equation

\[ I'(p) \;=\; J(p), \qquad\text{so}\qquad I(p) \;=\; \int J(p)\,dp \;+\; C, \]

and the constant \(C\) is pinned by evaluating both sides at a base point \(p_0\) where \(I(p_0)\) is known exactly — usually a value of the parameter at which the integrand vanishes identically, giving \(I(p_0)=0\).

For the opening example, with \(p=a\) and \(f=(x^a-1)/\log x\):

\[ I'(a)=\int_0^1 \frac{\partial}{\partial a}\frac{x^a-1}{\log x}\,dx =\int_0^1 x^a\,dx =\frac{1}{a+1}, \]

because \(\partial_a x^a = x^a\log x\) cancels the \(\log x\) in the denominator. Integrating, \(I(a)=\log(a+1)+C\), and the base point \(a=0\) makes the integrand identically zero, so \(I(0)=0\) forces \(C=0\). Hence \(\int_0^1 (x^a-1)/\log x\,dx=\log(1+a)\) — exactly Out[1] above. Mathilda even shows the first step for you; the \(a\)-derivative really is that simple:

In[1]:= D[(x^a - 1)/Log[x], a]
Out[1]= x^a

A little history

The rule is Leibniz's, and Euler used it heavily in the 18th century. Its 20th-century fame is due to Feynman, who learned it from Woods's Advanced Calculus as a schoolboy and later:

"So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's…" — R. P. Feynman

The uniform framing "every case is the first-order ODE \(I'(p)=J(p)\)" follows Boulnois (2023). A careful account of when the rule is valid — and the conditionally-convergent pitfalls where it is not — is Conrad's expository note.

References

  1. R. P. Feynman, Surely You're Joking, Mr. Feynman!, W. W. Norton, 1985 — "A different box of tools" (the popular account).
  2. F. S. Woods, Advanced Calculus, Ginn & Co., 1926 — the text Feynman learned the rule from.
  3. K. Conrad, "Differentiating under the integral sign," expository notes, kconrad.math.uconn.edu — rigorous hypotheses and worked examples (§12: conditional convergence).
  4. D. Boulnois, "A unified first-order ODE approach to differentiation under the integral sign," arXiv:2308.09619, 2023.
  5. G. B. Folland, Real Analysis, 2nd ed., Wiley, 1999 — the dominated- convergence justification (Thm 2.27).

2. How Mathilda realises it

The method runs the four steps above literally, and every step is symbolic — there is no numerical quadrature anywhere in it.

  1. Pick a parameter. The free symbols of the integrand other than \(x\) are the candidate parameters. Their signs are read from Assumptions.

  2. Differentiate and integrate. For a candidate \(p\), form \(g=\operatorname{Simplify}[\partial_p f]\) and evaluate the inner definite integral \(J=\int_a^b g\,dx\). If it does not close to a finite closed form, that parameter is abandoned and the next is tried.

  3. Integrate back over the parameter to get the antiderivative \(G(p)=\int J\,dp\).

  4. Fix the constant from a base point: \(I(p)=G(p)-G(p_0)+I(p_0)\), where \(p_0\) is a value at which \(f\) vanishes identically (\(I(p_0)=0\)) or the integral is directly computable.

  5. Verify. The result is accepted only if \(\operatorname{Simplify}[\,I'(p)-J\,]\equiv 0\) under the assumptions, together with the exact base value. The verification is a genuine proof, not a numeric spot-check — consistent with the project rule that Integrate performs no NIntegrate crosscheck.

Why the inner integrals use dedicated families

The one delicate part is step 2: the differentiated integral \(J\) is itself a parameter-dependent definite integral, and the general integrator is slow or hangs on exactly the half-line exponential/trigonometric forms Feynman's trick produces. So DiffUnderInt computes those inner integrals with its own closed-form family evaluators and never hands them to the general engine. The families it carries are:

Inner family Shape Closed form
Laplace / Fourier half-line \(\int_0^\infty x^n e^{\alpha x}\{1,\cos,\sin\}\,dx\), \(\operatorname{Re}\alpha<0\) \(n!\,/\,(-\alpha)^{n+1}\) combinations
sinc / Frullani \(\int_0^\infty (\cdots)/x\,dx\) via \(\int_0^\infty M(s)\,ds\), \(M\) the Laplace image
even-rational half-line \(\int_0^\infty P(x)/Q(x^2)\,dx\) \(v=x^2\Rightarrow\) Beta integrals
general (non-even) rational half-line \(\int_0^\infty R(s)\,ds\) real \(\operatorname{ArcTan}\)/\(\log\) boundary values
Gaussian moment \(\int_0^\infty x^n e^{-p x^2}\{1,\cos\}\,dx\) in \(\sqrt\pi\), \(e^{-q^2/4p}\)

The Gaussian parameter back-integration \(\int c\,e^{-k p^2}\,dp\) is supplied directly as an Erf (the engine produces no such antiderivative). A form outside every family is declined — the integral comes back unevaluated, fast, never a wrong value (see §7).

DiffUnderInt is tried near the end of the definite Integrate cascade, after the residue and Newton–Leibniz methods — it is the tool for integrals that are only tractable through a parameter.

3. Invoking the method

Pin the method with Method -> "DiffUnderInt" (the long alias "DifferentiationUnderIntegral" also works), or call the builtin directly as Integrate`DiffUnderInt[f, {x, a, b}]. All three are equivalent:

In[1]:= Integrate[(x^a - 1)/Log[x], {x, 0, 1}, Method -> "DifferentiationUnderIntegral"]
Out[1]= Log[1 + a]

In[2]:= Integrate`DiffUnderInt[(x^a - 1)/Log[x], {x, 0, 1}]
Out[2]= Log[1 + a]

The integrand must contain a free parameter to differentiate with respect to — an integral of numbers only is not this method's business. Supply Assumptions to pin the signs of the parameters; they fix the convergence gates and are used to clean the closed forms:

In[1]:= Integrate[Sin[a x]^2/x^2, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0]
Out[1]= 1/2 Pi a

You do not normally need to name the parameter: the method tries each free symbol in turn and keeps the first that yields a verified closed form.

4. Worked examples

Every transcript below is reproduced from the current build. Timings (from // Timing) are indicative only.

4.1 The logarithm trick

The archetype: a Log[x] in the denominator killed by \(\partial_a x^a = x^a\log x\). With two exponents the answers just superpose (linearity of the parameter integral):

In[1]:= Integrate[(x^a - 1)/Log[x], {x, 0, 1}, Method -> "DiffUnderInt"] // Timing
Out[1]= {0.005486, Log[1 + a]}

In[2]:= Integrate[(x^a - x^b)/Log[x], {x, 0, 1},
          Method -> "DiffUnderInt", Assumptions -> {a > -1, b > -1}]
Out[2]= Log[1 + a] - Log[1 + b]

4.2 Frullani integrals

A Frullani integral \(\int_0^\infty \frac{g(ax)-g(bx)}{x}\,dx\) is a natural target: differentiating in \(a\) removes the \(1/x\) and leaves a standard transform. The classic exponential and arctangent cases:

In[1]:= Integrate[(Exp[-a x] - Exp[-b x])/x, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> {a > 0, b > 0}]
Out[1]= -Log[a] + Log[b]

In[2]:= Integrate[(ArcTan[a x] - ArcTan[b x])/x, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> {a > 0, b > 0}]
Out[2]= 1/2 Pi (Log[a] - Log[b])

The first is the textbook Frullani value \(\log(b/a)\). Attaching an oscillatory factor \(\cos(cx)\) gives a two-parameter version that still closes:

In[1]:= Integrate[(Exp[-a x] - Exp[-b x])/x Cos[c x], {x, 0, Infinity},
          Method -> "DiffUnderInt",
          Assumptions -> {a > 0, b > 0, Element[c, Reals]}]
Out[1]= -1/2 Log[a^2 + c^2] + 1/2 Log[b^2 + c^2]

i.e. \(\tfrac12\log\!\big((b^2+c^2)/(a^2+c^2)\big)\). The split-Log form is the tidy real result of a Laplace-transform inner integral.

4.3 Sinc and Dirichlet integrals

Integrands with \(1/x^2\) or \(1/x\) reduce, after one differentiation, to a sinc integral over the half-line. The two standard \(\pi a/2\) results:

In[1]:= Integrate[Sin[a x]^2/x^2, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0] // Timing
Out[1]= {0.063488, 1/2 Pi a}

In[2]:= Integrate[(1 - Cos[a x])/x^2, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0]
Out[2]= 1/2 Pi a

Adding an exponential window differentiates to \(\int_0^\infty e^{-ax}\cos(bx)\,dx\), whose parameter integral is the arctangent:

In[1]:= Integrate[Exp[-a x] Sin[b x]/x, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> {a > 0, b > 0}]
Out[1]= 1/2 Pi - ArcTan[a/b]

which is \(\operatorname{arctan}(b/a)\) (the two differ by the identity \(\operatorname{arctan} t+\operatorname{arctan}(1/t)=\pi/2\)).

4.4 Rational half-line with a logarithm

Here the parameter sits inside a logarithm; differentiating turns it into a rational function integrated against \(1/(1+x^2)\):

In[1]:= Integrate[Log[1 + a^2 x^2]/(1 + x^2), {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0]
Out[1]= Pi Log[1 + a]

In[2]:= Integrate[Log[1 + a^2 x^2]/(x^2 (1 + x^2)), {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0] // Timing
Out[2]= {0.050109, Pi (a - Log[1 + a])}

In[3]:= Integrate[ArcTan[a x]/(x (1 + x^2)), {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> a > 0]
Out[3]= 1/2 Pi Log[1 + a]

4.5 A decaying sinc — the non-even rational family

Combining an exponential decay with a \((1-\cos ax)/x^2\) window differentiates to a decaying sinc \(\int_0^\infty e^{-cx}\sin(ax)/x\,dx=\operatorname{arctan}(a/c)\). Its Laplace image \(M(s)=a/((s+c)^2+a^2)\) is not even in \(s\), so Mathilda integrates it with the general non-even rational half-line evaluator, which returns a real arctangent directly:

In[1]:= Integrate[Exp[-c x] (1 - Cos[a x])/x^2, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> {a > 0, c > 0}] // Timing
Out[1]= {0.653151, a ArcTan[a/c] - 1/2 c (-2 Log[c] + Log[a^2 + c^2])}

that is, \(a\operatorname{arctan}(a/c)-\tfrac{c}{2}\log\!\big(1+a^2/c^2\big)\).

4.6 A Gaussian integral

A Gaussian weight is no obstacle. Differentiating \(e^{-x^2}\sin(ax)/x\) in \(a\) removes the \(1/x\) and leaves the cosine moment \(\int_0^\infty e^{-x^2}\cos(ax)\,dx=\tfrac{\sqrt\pi}{2}e^{-a^2/4}\); integrating that back over \(a\) produces an error function:

In[1]:= Integrate[Exp[-x^2] Sin[a x]/x, {x, 0, Infinity},
          Method -> "DiffUnderInt"] // Timing
Out[1]= {0.15825, 1/2 Pi Erf[1/2 a]}

The Erf comes from the Gaussian parameter back-integration \(\int \tfrac{\sqrt\pi}{2}e^{-a^2/4}\,da\), which DiffUnderInt supplies in closed form because the general integrator does not produce it.

5. Verification and the base point

Because the answer is reconstructed from a derivative, DiffUnderInt proves each result before returning it. Two things are checked:

  • The differential equation. \(\operatorname{Simplify}[\,I'(p)-J\,]\) must reduce to the literal 0 under your Assumptions. This is a symbolic identity check on elementary functions — a genuine proof, not a sampled comparison. (PossibleZeroQ is deliberately not used here: its shrinkage heuristic false-positives on decaying expressions, which could mis-accept a wrong result.)

  • The base value. The additive constant is fixed from an exact \(I(p_0)\), most often a parameter value at which the integrand is identically zero, so \(I(p_0)=0\) with no computation at all.

Together these make the closed form correct by construction. The conditionally-convergent pitfall that makes naive differentiation under the integral sign fail (Conrad, §12) is caught automatically: a non-integrable \(\partial_p f\) simply fails to close, and that parameter is skipped rather than producing a spurious value.

6. The families at a glance

The inner definite integral \(J\) is matched against the closed-form families listed in §2. Their combined reach is: half-line integrals of poly × exp × {1, sin, cos} (with or without a \(1/x^k\) window), even and non-even rational functions, and Gaussian moments — plus the parameter-side integrals of the arctangents, logarithms, rationals, and Gaussians those produce. That is enough to close the standard corpus of "Feynman's trick" exercises, as §4 shows.

7. Limitations

  • A free parameter is required. With no symbol to differentiate against, the method does not apply and returns the input unevaluated. Reach for another method in the cascade ("NewtonLeibniz", "Residue", "Mellin").

  • Piecewise results are declined. When the true answer is a genuine case-split — for example \(\int_0^\infty \frac{\sin(ax)\sin(bx)}{x^2}\,dx = \frac{\pi}{2}\min(a,b)\) — the differentiated inner integral is itself conditional, which the families do not carry. The integral is returned unevaluated, cleanly and fast:

In[1]:= Integrate[Sin[a x] Sin[b x]/x^2, {x, 0, Infinity},
          Method -> "DiffUnderInt", Assumptions -> {a > 0, b > 0}]
Out[1]= Integrate[(Sin[a x] Sin[b x])/x^2, {x, 0, Infinity},
          Assumptions -> {a > 0, b > 0}, Method -> "DiffUnderInt"]

  • Some inner integrals are out of family. Finite-period trigonometric integrals over \(\{0,\pi\}\), finite-interval radicals, and half-line log × rational base values fall outside the current evaluators and are declined rather than guessed. The purely oscillatory Sine–Gaussian moment \(\int_0^\infty e^{-x^2}\sin(ax)\,dx\) (a Dawson/Erfi form) is likewise not carried — but, as §4.6 shows, the cases that differentiate into a cosine moment still close.

  • No numeric fallback. Consistent with the rest of Integrate, the method never approximates: it either returns a proven closed form or the unevaluated input.

See also