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guys this is the question i come across- to find the value of

$$\int_{-∞}^{∞}e^{-2t} \delta'(t).dt$$

where $ \delta(t) $ is the usual delta function or impulse function. The problem is solved as the given integral is equal to $$ -\int_{-∞}^∞ \frac{d}{dt} e^{-2t}. \delta(t).dt $$ $$= 2e^{-2t}|_{t=0} $$ $$=2$$ Can someone please explain how this is solved?

RonKhuman
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1 Answers1

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It's not an interchange of differentiation and integration. It's integration by parts.

joriki
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  • Can you please elaborate on how the first term reduces to zero? i am at a lost here. thanks – RonKhuman Jun 23 '18 at 16:04
  • @HarikishanChandam: Quite generally, when you're integrating over the entire real line and the integrand includes delta distributions or their derivatives, the boundary terms at infinity vanish, since the delta distributions only care about the local behaviour of their test functions. If you think of the delta distribution as a sort of limit of regular functions, then those functions tend to zero at infinity. – joriki Jun 23 '18 at 16:09
  • thanks again.,if i use the integration by parts, then the first term becomes (e^-2t )* delta(t) with values ∞ and -∞ to be substituted. won't one value become an indeterminate form of 0x∞ ? – RonKhuman Jun 23 '18 at 16:15
  • @HarikishanChandam: Ah, interesting. I hadn't thought of it that way. You're actually right that this is problematic, in that if you treat the delta distribution rigorously, then the test functions you can integrate them with are usually required to be Schwartz functions, and the exponential wouldn't be allowed. See also this answer on math.SE. As you can see there, it may depend on the context of what you're actually trying to do here. – joriki Jun 23 '18 at 16:41