Does this hold in general? $$ \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty} f(x,t) \mathrm{d}x = \int_{-\infty}^{\infty}\frac{\partial}{\partial t}f(x,t) \mathrm{d}x. $$ I know it is true if the bounds on the integral are finite but can the result be extended to improper integrals? Also if it is true in special cases, what are those cases?

Thanks a lot!

  • $\begingroup$ See this $\endgroup$ – Hayden Feb 20 '17 at 4:19
  • $\begingroup$ The dominated convergence theorem implies it should work if there is an integrable function $\int_{-\infty}^{\infty} |M(x)| \, \mathrm{d}x < \infty$ such that $|\partial_t f(x,t)| \le |M(x)|$ for all $x$ and all $t$. Is that the kind of statement you are looking for $\endgroup$ – user399601 Feb 20 '17 at 4:22
  • $\begingroup$ @Hayden That reference does not apply to improper integrals. $\endgroup$ – Mark Viola Feb 20 '17 at 5:35
  • 1
    $\begingroup$ @user399601 Since the OP has explicitly asked about the results for improper integrals (and tagged improper integrals), I speculate bravely that Lebesgue Theory is likely off the table. $\endgroup$ – Mark Viola Feb 20 '17 at 5:37

No, the equality does not hold in general.


For a first example, the integral $I(x)$ as given by


converges uniformly for all $|x|\ge \delta>0$. But the integral of the derivative with respect to $x$, $\int_{-\infty}^\infty \cos(xt)\,dt$ diverges for all $x$.


As another example, let $J(x)$ be the integral given by

$$J(x)=\int_0^\infty x^3e^{-x^2t}\,dt$$

Obviously, $J(x)=x$ for all $x$ and hence $J'(x)=1$. However,

$$\int_0^\infty (3x^2-2x^4t)e^{-x^2t}\,dt=\begin{cases}1&,x\ne 0\\\\0&,x=0\end{cases}$$

Thus, formal differentiation under the integral sign leads to an incorrect result for $x=0$ even though all integrals involved are absolutely convergent.

Sufficient Conditions for Differentiating Under the Integral

If $f(x,t)$ and $\frac{\partial f(x,t)}{\partial x}$ are continuous for all $x\in [a,b]$ and $t\in \mathbb{R}$, and if $\int_{-\infty}^\infty f(x,t)\,dt$ converges for some $x_0\in[a,b]$ and $\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$ converges uniformly for all $x\in [a,b]$, then

$$\frac{d}{dx}\int_{-\infty}^\infty f(x,t)\,dt=\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$$

  • $\begingroup$ Did you mean $\delta \gt 0$ in first example? $\endgroup$ – Error 404 Feb 20 '17 at 5:42
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    $\begingroup$ @VikrantDesai Yes, of course. I've edited the typo. Thank you for the catch! $\endgroup$ – Mark Viola Feb 20 '17 at 5:42

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