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I want to evaluate the following integral integral of ln(dx+1), yes the integral is inside the ln, NOT outside

The inspiration came from this video integral of x^dx-1, where the integral was divided and then multiplied by dx, so when evaluating my integral I wanted to use the same trick to get this integral. Then I took the limit as dx approaches 0 to get infinity. So finally I ended up with an integral of infinity * dx which is just equal to infinity.

I know the question doesn't make any mathematical sense, so I'm just asking if the question can be solved "symbolically"? Am I doing any mistakes in my calculation of the inegral? Is there another way to actually solve it?

Many thanks,

zcxqwezxc
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3 Answers3

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Let our putative integral be $I$: $$I=\int_a^b\ln(1+dx)$$ Exponentiating both sides turns the RHS into a type I product integral: $$e^I=\prod_a^b(1+dx)=\exp\left(\int_a^b1\,dx\right)=e^{b-a}$$ Hence we assign the value $b-a$ to $I$.

Parcly Taxel
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I would just say $$ \log(1+dx) = dx + O(dx^2), $$ so $$ \int \log(1+dx) = \int dx = x + C. $$

mjqxxxx
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  • My attempt is to justify this, what I called in my answer the Taylor series formula, because many people are unhappy with treating $dx$ like an algebraic object. – Ninad Munshi Apr 11 '21 at 05:24
  • I would think of $dx$ as a particular value (not a function of $x$) tending to zero. This is consistent with the Riemann or Lebesgue integral. In which case (though as you say, it doesn't really "make mathematical sense"), you can always just work with the first-order Taylor expansion of any function of $dx$, ignoring any terms that are $o(dx)$. – mjqxxxx Apr 11 '21 at 16:51
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You can always assign strange expressions the value from an associated Riemann sum and rigorously ask if the summation converges to what we would naively expect from using the Taylor series formulas

$$ \int_a^b\ln(dx+1) \mathop{``="}\lim_{n\to\infty}\sum_{i=1}^n \ln\left(\frac{b-a}{n}+1\right) = \lim_{n\to\infty} n\ln\left(\frac{b-a}{n}+1\right) = \ln e^{b-a} = b-a$$

This will work for any "weird" integral expression if we are careful about correctly reassigning limit terms.

user10354138
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Ninad Munshi
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  • Can you please explain 2nd to 3rd step ? – Koro Apr 11 '21 at 05:17
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    @Koro the summand is a constant w.r.t. the summation variable. – Ninad Munshi Apr 11 '21 at 05:17
  • @Mickael I am confused by your comments. The summation is not infinite, there are only $n$ terms, as is the case with all Riemann sums. – Ninad Munshi Apr 11 '21 at 05:22
  • @Mickael firstly it is a summation from 1 to $n$. Second, as I already explained in the earlier comment, the summand is a constant. What is the sum of a constant? The summand times the number of terms. – Ninad Munshi Apr 11 '21 at 05:35
  • @Mickael ......how are they different? – Ninad Munshi Apr 11 '21 at 05:39
  • @Mickael I don't know if it's because English is not your native language (it isn't mine either) but you need to read their answer more carefully instead of asking other people questions. What did they say the value of $I$ was at the end of their post? – Ninad Munshi Apr 11 '21 at 05:41