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Recently I was studing the Feynman integration technique (differentiation under the integral sign), but I allways get stuck on the general function definition, and I was wondering if there is some kind of formula, trick, table, or something else that can help me evaluate such a funtion.

Thankfully liuzp.

Update:

For functions like: $$\int_a^b \frac{f(x)}{g(x)} \hspace{1cm} f(x,\alpha )=e^{\alpha\ g(x) + c(x)} + d(x)$$ where $c(x)$ and $d(x)$ are special functions that make $f(x, \alpha ) = f(x)$ for some special value of $\alpha$.

But I've also founf that this won't work for all cases, and for cases where you haven't a denominator.

This also have a limitation because $g(x) \in \{\mathbb{P}^2,\ln[h(x)]\} $

liuzp
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  • Do you mean Feynman's "differentiation under the integral" technique? Or perhaps this? Where exactly are you stuck? – Robert Israel Nov 13 '18 at 19:43
  • Yes, differentiation under the integral sign. I forgot to add in the description. I'm stuk on fiding the general function given the especfic case, for which you wnat to solve. – liuzp Nov 13 '18 at 21:21

1 Answers1

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Here's a typical example.

$$ J = \int_0^\infty \frac{\arctan(x)}{x(1+x^2)}$$

We'd like to get rid of that nasty arctan. We know the derivative of $\arctan$ is nice. So we generalize to

$$ J(t) = \int_0^\infty \frac{\arctan(tx)}{x(1+x^2)}$$

take the derivative with respect to $t$, and the rest is easy...

Robert Israel
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  • OK, you have just increased the sise of my table, but there is no easy way to do this? I can't see a patern emerging from all of this, when I get $x \cot(x)$ I should use $\arctan(\alpha \tan(x)) \cot(x)$, but when I get something else I need to do a nasty substitution all of the sudently. – liuzp Nov 14 '18 at 00:45
  • I've updated the post, could you give a new look to it? – liuzp Nov 16 '18 at 02:37