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Let $I_n (a)$= $\int_0^1 x^a (\ln(x))^n \,dx$ for $a > 0$ and $n=0,1,2,...$

Find $I_0 (a)$. Differentiate in $a$ to find $I_n (a)$ for all $n \ge 1$

The first part, I'm assuming I can sub in $n=0$ so I have to integrate $x^a$ which is clear, and then sub in the values.

But for the second part, I have little idea. From a google search, there is something to do with differentiating under the integral, but I have no idea how to do this or how to apply it. Any advice appreciated!

clathratus
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User2121
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  • Just apply integration by parts differentiating $(\ln{(x)})^n$ and integrating $x^a$. What do you get? – Peter Foreman Nov 24 '19 at 13:48
  • For $a<1$ we have a problem... $a=1/2$ and n=2 the derivative $\frac{-1}{2}x^{-1/2}(\log{x})^2$ is not integrable..the integral of the derivative diverges..are you sure that $a>0$ and not $a \geq 1$?? – Marios Gretsas Nov 24 '19 at 13:51
  • Well, you already know how to solve it, just differentiate under the integral sign with respect to a. See also here:: https://math.stackexchange.com/a/3372053/515527 – Zacky Nov 24 '19 at 14:25

1 Answers1

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If we perform one round of integration by parts we can start to see a pattern:

$$ I_n(a)= \int_0^1 x^a\ln^n(x)dx=\left . \frac{1}{a+1}x^{a+1}\ln^n(x)\right|_0^1 -\frac{1}{a+1}\int_0^1 n x^a\ln^{n-1}(x)dx =-\frac{n}{a+1}I_{n-1}(a). $$

From here we can prove by induction that

$$ I_n(a) = (-1)^n\frac{n!}{(a+1)^{n+1}}. $$

The base case for $n=0$ is simple since $\int_0^1x^adx=\frac{1}{a+1}$. Assuming it’s true for $I_n(a)$ then we can use the computation in the above line to show that

$$ I_{n+1}(a)= -\frac{n+1}{(a+1)}I_n(a)=(-1)^{n+1}\frac{(n+1)!}{(a+1)^{n+2}}. $$

Mnifldz
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  • Thanks. My only query is how you get to the line after "From here we can prove by induction that"...? – User2121 Nov 25 '19 at 01:02
  • Try looking up proof by induction. The way these statements are proved are of the form “show that the statement works for $n=0$, then show that if it’s true for $n$ then it also must be true for $n+1$.” The way I solved it was by inferring a pattern from the first relation, and then using induction to show that it is in fact true for all $n$. – Mnifldz Nov 25 '19 at 01:05
  • I know about induction and how to use it, I'm just unsure why you've let $$I_n(a) = (-1)^n\frac{n!}{(a+1)^{n+1}}.$$ – User2121 Nov 25 '19 at 01:08
  • I didn’t “let” $I_n(a)$ equal anything. I made a claim as to its value, and then proved that claim true using induction. – Mnifldz Nov 25 '19 at 02:27
  • Ok, I get that but that value comes out of the blue for me. Where has it come from? – User2121 Nov 25 '19 at 02:37
  • It comes from deriving the recursion relation between $I_n(a)$ and $I_{n-1}(a)$. There is a multiplicative factor of $-\frac{n}{a+1}$ at the $n$th level. If we can figure out $I_0(a)$, which is pretty easy to find, we can the figure out $I_1(a)=-\frac{1}{a+1}I_0(a)=-\frac{1}{(a+1)^2}$, and all other values of $I_n(a)$ in this fashion. Jumping to the expression I gave is really based on this, but positing that value without offering a proof isn’t proper. Induction is the correct way to justify this. At least it’s quicker than integrating by parts the whole way. – Mnifldz Nov 25 '19 at 04:13