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Say you have the integral $$g(t)=\int e^{-t}f(t)dt$$

Is it possible to rewrite this in some way such that we can find $g(t)$ for any $f$, so long as $f$ is itself tractable?

E.g. is there a way to rewrite it as $g(t)= A(t)+ B(t)\int f(t)dt$ for tractable A and B?

user56834
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  • If the formula you covet existed, one would have $$g=A-A'(B/B')+(h-B)(B/B')f$$ where $h(t)=e^{-t}$, for every $f$. In particular, $f=0$ would yield $g=0$ hence $A-A'(B/B')=0$, and you would be left with $g=mf$ for some function $m$ independent of $f$, which is clearly impossible. – Did Apr 26 '17 at 08:20
  • The expression $\int e^{-t} f(t) \text{d}t$ does not depend on $t$. Do you perhaps mean $g(t) = \int_{0}^{t} e^{-s} f(s) \text{d}s$. Or maybe $g(t) = \int_{\mathbb{R}} e^{-s} f(t-s) \text{d}s$? – Peter Apr 26 '17 at 09:02
  • As a rule of thumb, when you add a factor to an integrand, it becomes less tractable. –  Apr 26 '17 at 14:08

1 Answers1

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The answer is "no".

You can find examples where $\int f(t)\;dt$ is easy, but $\int e^{-t} f(t)\,dt$ is difficult or impossible in elementary functions. For example: \begin{align} &\int\frac{dt}{t} = \log t\qquad\text{is elementary, but} \\ &\int\frac{e^{-t}}{t}\;dt\qquad\text{is an "exponential integral", not elementary} \end{align}

Because of this, we have tables of "Laplace transforms" to look up such things when they are known.

GEdgar
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