When to differentiate under the integral sign? I'm finishing up a semester of multivariable calculus and will be taking a course on analysis this Spring. In any of the calculus courses I've taken, we never covered anything beyond the standard techniques of integration ($u$-sub, parts, etc.) One of the techniques I saw used recently which I had not heard of was differentiation under the integral sign, which makes use of the fact that:
$$\frac{d}{dx} \int_a^bf(x,t)dt = \int_a^b \frac{\partial}{\partial x}f(x,t)dt $$
in solving integrals. My question is, is there ever an indication that this should be used? Is there any explainable intuition or rule of thumb for the use of differentiation under the integral sign?
 A: It usually comes up when you are dealing with functions defined in terms of an integral but also can be used to clean up ugly integrals by introducing a new parameter and differentiating with respect to said new parameter. Look up Feynman integration, there are lots of instructive videos and examples of this technique.
A: It can require more originality than other methods. If you work through the linked examples, you might develop an instinct for when to try it. But I don't think there's a broad rule for when to try it. However, when you see an integrand of the form $f(x)x^s\ln^nx$ with $n\in\Bbb N$ where the problem would be easy for $n=0$, that's a good sign. So is an integrand of the form $h(x)f(1+ag(x))/g^\prime(x)$.
A: Differentiation under integral signs, better known as the Feynman’s trick, is not a standard integration technique taught in curriculum calculus. Nevertheless, it is widely utilized outside classrooms and may appear somewhat magic to those seeing it the first time.
Despite the mystique around it, it is actually rooted in double integrals. A bare-bone illustrative example is
$$I=\int_0^1\int_0^1 x^t dt \>dx= \ln2$$
The natural approach is to integrate $x$ first and then $t$. But, an unsuspecting person may integrate $t$ first and then encounter
$$I=\int_0^1\frac{x-1}{\ln x} dx$$
Now, he/she is stuck since there is not easy way out. Fortunately, there is, which is to differentiate $J(t)$ below under the integral, i.e.
$$J(t)=\int_0^1\frac{x^t-1}{\ln x} dx,\>\>J(t)' = \int_0^1 x^t dx= \frac{1}{1+t}
\implies I=\int_0^1 J(t)'dt=\ln 2$$
A knowledgeable math person, aware of its double-integral origin, would just undo the $t$-integral to reintroduce the double form, and then integrate in the right order,
$$I=\int_0^1\frac{x-1}{\ln x} dx=\int_0^1\int_0^1 x^t dt dx = \int_0^1 \frac1{t+1}dt= \ln 2$$
The two approaches are in fact equivalent, with the double-integrals actually more straightforward. The Feynman’s trick is appealing, since it ‘decouples’ a double-integral in appearance, especially when the embedded double-integral is not immediately discernible. When a seemingly difficult integral is encountered, the differentiation trick is often employed to transform the original integrand to a manageable one.
As a practical example, the trick can be used for deriving the well-known integral
$$I=\int_0^\infty \frac{\sin x}x dx=\frac\pi2$$
with $J(t)=\int_0^\infty \frac{\sin x}x e^{-tx}dx$, $
J’(t)=-\frac{1}{1+t^2}$ and $I=- \int_0^\infty J’(t)dt$.
A: if you integrate a "differentiation" function, it means you are trying to get the original value. in most of the questions you have to integrate the original value 
