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Is integration by parts the best method to solve the integral $\displaystyle \int_{0}^{1} \ln(x)dx$? or should i rather use the Feynman differentiation under the integral sign trick?

Dave L. Renfro
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user1729
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    What qualifies one method as better than another? – Sassatelli Giulio Sep 18 '22 at 17:46
  • Integration by parts is nice in that you can apply it off the bat, and it works nicely for pretty much any $\int f(x) dx$ for $f$ the inverse of a well-known and differentiable function. Feynman integration probably can work for a large class of integrals, but the issue is setting up the parameterization needed; I don't know if there are general rules you should follow for that, though. I guess $\int_0^1 \ln(ax) , dx$ would be the setup? But especially for more complicated functions, it's hard to say. It's basically a tradeoff of "I know this will work, even if longer" versus "this is cooler" – PrincessEev Sep 18 '22 at 17:46
  • Though I suppose that, since Feynman's trick basically amounts to making and solving a differential equation, in some sense it's longer, so I find it hard to think of a way in which Feynman integration might be "better" here, other than it looking cooler. – PrincessEev Sep 18 '22 at 17:48
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    @PrincessEev I think the OP may have had $\left.\partial_a\int_0^1x^{a-1}dx\right|_{a=1}$ in mind. Anyway, substituting $y=-\ln x$ is yet another option. – J.G. Sep 18 '22 at 17:49
  • @J.G. That should perhaps be called "the replica trick". – user619894 Sep 18 '22 at 17:57
  • The classical simple method : write your integral $\int_0^1 u v' dx$ with $u=\log x$ and $v=x$, then apply integration by parts... – Jean Marie Sep 18 '22 at 18:03
  • @user619894 As in this? I guess it's both. – J.G. Sep 18 '22 at 18:20
  • Why not use $\int_0^1 \ln xdx = -\int_\infty^0 e^xdx = -1$? – Théophile Sep 21 '22 at 21:49

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I think the most standard way to solve it would be to just use parts and Calc II strategies. Leibniz rule seems just overkill.

$$\int^1_0\ln(x)\cdot 1\text{ d}x=\ln(x)\cdot x\Big|^1_0-\int^1_0\frac{1}{x}\cdot x\text{ d}x=(x\ln(x)-x)\Big|^1_0=-1$$

Captain Chicky
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