# How to solve combinations of differentials and integrals?

In the first chapter of Nearing's book "Mathematical tools for physics" (available online) I encountered an interesting combination of differentials and integrals - which I don't fully understand:

(a) $$\frac{\mathrm{d} }{\mathrm{d}\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2}dx=-\int_{-\infty}^{\infty}x^2e^{-\alpha x^2}dx$$

(b) $$\frac{\mathrm{d} }{\mathrm{d} x}\int_{0}^{x}e^{-x t^2}dt=e^{-x^{3}}-\int_{0}^{x}t^2e^{-x t^2}dt$$

(c) $$\frac{\mathrm{d} }{\mathrm{d} x}\int_{x^2}^{\,\sin x}e^{x t^2}dt=e^{x\, \sin^2 x} \,\cos x-e^{x^{5}}2x+\int_{x^2}^{\,\sin x}t^2e^{x t^2}dt$$

I can't see in which order you have to do which rules to arrive at the solutions. Could anyone please give me the steps in between? Thank you!

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It's a Leibniz integral rule -see e.g. in wikipedia by the link.

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I think you can also prove the first statement just by applying the definition of the derivative to the integral in question (where you then need to justify moving $lim_{h\rightarrow 0}$ from outside the integral to inside). –  user3180 Apr 11 '11 at 7:09
I will say that for the integral over a non-compact set you should be sure that some conditions are satisfied. Otherwise you cannot say that your result is correct. –  Ilya Apr 11 '11 at 7:11
@Gortaur: I fixed the link, there was a superfluous / at the end. Could you explain to me what "bu the link" means? –  t.b. Apr 11 '11 at 7:27
@Theo: it was a misprint. I hope "bu" is not a slang. –  Ilya Apr 11 '11 at 7:39
Don't worry, it isn't. I thought it might be some sort of abbreviation and couldn't figure out what it might stand for. Thanks for the clarification. –  t.b. Apr 11 '11 at 7:43

Basically all those identities are just using differentiation under the integral sign. For example for the first one, you have a function of two variables $f(x, \alpha) : = e^{-\alpha x^2}$ and then the identity is obtained by interchanging the order of the derivative with the integral

$$\frac{\mathrm d}{ \mathrm d \alpha} \int_{-\infty}^{\infty} e^{-\alpha x^2} \, \mathrm{dx} =\frac{\mathrm d}{ \mathrm d \alpha} \int_{-\infty}^{\infty} f(x, \alpha) \, \mathrm{dx} = \int_{-\infty}^{\infty} \frac{\partial f(x, \alpha)}{\partial \alpha} \, \mathrm dx$$

$$= \int_{-\infty}^{\infty} \frac{\partial}{\partial \alpha} (e^{-\alpha x^2}) \, \mathrm dx = \int_{-\infty}^{\infty} -x^2 e^{-\alpha x^2} \, \mathrm{dx}$$

Of course some conditions must be satisfied for this in order to work, but you can check what those conditions are in the previous link. You can find this treated in most analysis books. For example it is treated in chapter 9 of Rudin's Principles of Mathematical Analysis under the section "Differentiation of Integrals".

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Thank you! Why do you have an extra term in (b) and two extra terms in (c)? Because you have to differentiate acc. to the boundaries of the integral? But in (a) you also have boundaries (- to + infinity)??? –  vonjd Apr 11 '11 at 8:04
@vonjd The difference is that you have a variable in the limits of the integrals in parts b) and c) so you also need to apply the Fundamental Theorem of Calculus with the chain rule when differentiating. –  Adrián Barquero Apr 11 '11 at 8:50