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Let$\ I_n = \int_{0}^{1} \frac{1}{(1+x^2)^n} dx$. Prove that for every positive integer n, $$\ 2nI_{n+1} = 2^{-n}+(2n-1)I_n$$

Attempt

$$\ I_{n+1} = \int_{0}^{1} \frac{1}{(1+x^2)^{n+1}} dx$$ $$\ I_{n+1} = \int_{0}^{1} \frac{1+x^2}{(1+x^2)^{n+1}} dx-\int_{0}^{1} \frac{x^2}{(1+x^2)^{n+1}} dx$$ $$\ I_{n+1} = I_n-\int_{0}^{1} \frac{x^2}{(1+x^2)^{n+1}} dx$$ $$\ I_{n+1} = I_n-[\frac{-2^{-n}}{2n}+\frac{1}{n}\int_{0}^{1} \frac{1}{(1+x^2)^{n}} dx]$$ $$\ 2nI_{n+1} = 2nI_n+2^{-n}-2I_n $$ $$\ 2nI_{n+1} = (2n-2)I_n+2^{-n} $$

Can somebody please tell me the mistake I am doing?

mathnoob123
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1 Answers1

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Can somebody please tell me the mistake I am doing?

I see a mistake when you integrate by parts, you have forgotten the factor $\dfrac12$ in front of the second integral, one rather has, for $n\ge1$, $$ \int_{0}^{1} \frac{x^2}{(1+x^2)^{n+1}} dx=\left[x\cdot-\frac{1}{2n(1+x^2)^{n}}\right]_{0}^{1}+\frac1{2n}\int_{0}^{1} \frac{1}{(1+x^2)^n} dx. $$ from which $$ I_n-I_{n+1}=-\frac{2^{-n}}{2n}+\frac1{2n}I_n $$ or

$$\ 2nI_{n+1} = 2^{-n}+(2n-1)I_n$$

as announced.

Olivier Oloa
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  • How is there a factor of 1/2 present in thee second integral. The coefficient due to integrated term is 1/2n and coefficient due to differentiated term is 2. Thus both 2's cancel out. – mathnoob123 Mar 15 '17 at 11:04
  • Wait are we first going to divide x^2 by x( from the integrated part) or will we first differentiate x^2 and then divide the X? – mathnoob123 Mar 15 '17 at 12:43
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    We use $\int u dv=uv-\int v du$ with $u=\frac x2$ and $dv=\frac{2x}{(1+x^2)^{n+1}}$ giving $du=\frac12$ and $v=-\frac{1}{n(1+x^2)^{n}}$. Tell me if it is Ok. – Olivier Oloa Mar 15 '17 at 13:16
  • Can you tell me what will be v and du if $\ dv = \frac{1}{(1+x^2)^{n+1}}$ and $\ u = x^2 $? – mathnoob123 Mar 15 '17 at 14:40
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    @FaiqRaees If we knew an antiderivative of $dv = \frac{1}{(1+x^2)^{n+1}}$ we would directly use it to find out what $I_n$ is... But this is not the case, that's why we rather take $dv = \frac{2x}{(1+x^2)^{n+1}}$ which is way easier to integrate... – Olivier Oloa Mar 15 '17 at 14:50
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    Oh okay I was mistaken by considering that $\ \int f(x)^n =\frac{f(x)^{n+1}}{(n+1)*\frac{d}{dx}f(x)} $ – mathnoob123 Mar 15 '17 at 14:53