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Why is approach 2 wrong here? I am relatively newe to calculus please explain in detail what is going wrong... Evaluate: $ F(\alpha) = \int_0^1 \frac{x^\alpha-1}{\ln x} dx $

Approach 1: (Correct)

Using Newton Leibniz Theorem, \begin{align} F'(\alpha) &= \int_0^1 \frac{\delta}{\delta\alpha} \left(\frac{x^\alpha-1}{\ln x}\right) dx \\ \implies F'(\alpha) &= \int_0^1 x^{\alpha} dx\\ \implies F'(\alpha) &= \frac{1}{\alpha+1} \\ \implies F(\alpha) &= \ln (\alpha+1) + c \end{align}

We can get $ c=0 $ from $ F(0)=0 $, Hence $$ F(\alpha) = \ln (\alpha+1)$$

Approach 2: (Incorrect) $$F(\alpha) = \int_0^1 \frac{x^\alpha}{\ln x} dx - \int_0^1 \frac{1}{\ln x} dx $$ Substituite $ x^{\alpha+1}=t $ $$ F(\alpha) = \int_0^1 \frac{1}{\ln t} dt - \int_0^1 \frac{1}{\ln x} dx $$ $$ F(\alpha) = 0 $$

cjferes
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Akshit Chhabra
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  • The integral is valid near the right endpoint in the limit only. You can't exchange the integral of the difference with the difference of the integrals when that right end limit does not exist. – Gregory Sep 27 '23 at 15:50
  • Is that the only mistake here? When my teacher was solving this question, he mentioned something about why the difference of integrals won't be 0 due to a non linear transform and hence we can't use this approach but I didn't get exactly why this was wrong... – Akshit Chhabra Sep 27 '23 at 15:57
  • I think if your integral ended at a value less than 1, you would probably see that they are both ok. – Gregory Sep 27 '23 at 15:58
  • @Gregory I don't see how approach 2 would work with a value less than 1 either... For example is we take integration 0 to 0.5 wouldn't approach 2 still give 0 as the answer? – Akshit Chhabra Sep 27 '23 at 16:01
  • No. Your limits of integration would have changed. – Gregory Sep 27 '23 at 16:04
  • @Gregory right. I totally missed that. Thanks for helping out – Akshit Chhabra Sep 27 '23 at 16:26
  • @TedShifrin it was $ F'(\alpha) = \frac{1}{\alpha+1} \implies F(\alpha) = \ln (\alpha+1) + C $ I simply integrated both sides with respect to $ \alpha $... I don't get what you're trying to say... I'm new to calculus so am not well versed with all the terminology etc. Also I didn't miss out on $ (\alpha+1)^2 $ in the denominator $ (\alpha+1) $ in both numerator and denominator and will get cancelled... I am sorry for not showing all the steps as I was in a hurry – Akshit Chhabra Sep 27 '23 at 16:29
  • @TedShifrin I just checked and I don't see why there would be two of them in the denominator $ x^\alpha dx = \frac{dt}{\alpha+1} $ will produce one in the denominator and $ \ln x = \frac{1}{\alpha + 1} \ln t $ will produce on in the numerator... Also I am sorry I didn't know i had that type and someone edited it for me... – Akshit Chhabra Sep 27 '23 at 16:43
  • @ТymaGaidash٠ I think I got your point about splitting the integral but how did you get that using the substitution? I'm not getting the same integral.. where did the ln x go? – Akshit Chhabra Sep 27 '23 at 16:48
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    @AkshitChhabra There was a differentiation error. We should get $\int_0^1\frac{1-t^{\frac1{a+1}-1}}{\ln(t)}dt$ – Тyma Gaidash Sep 27 '23 at 16:51
  • @ТymaGaidash٠ yes, that's right, thank you for helping me. – Akshit Chhabra Sep 27 '23 at 16:59
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    Something that is often overlooked in Calculus is that to apply a theorem like Newton-Leibnitz you need to make sure that its hypotheses are fulfilled. – Martin Argerami Sep 28 '23 at 05:50
  • The theorem you fatally misapplied was not Newton-Leibnitz (though it doesn't apply directly here since $F$ is improper at both its endpoints, this can be worked around), but rather the theorem $\int (f+g) = \int f + \int g$. That theorem only holds if at least one of $\int f$ or $\int g$ is finite. – Paul Sinclair Sep 28 '23 at 17:57

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