I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove. $$I_n=\int\frac{\mathrm{d}x}{(ax^2+b)^n}\\I_n=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$ I tried the substitution $x=\sqrt{\frac ba}t$, and it gave me $$I_n=\frac{b^{1/2-n}}{a^{1/2}}\int\frac{\mathrm{d}t}{(t^2+1)^n}$$ To which I applied $t=\tan u$: $$I_n=\frac{b^{1/2-n}}{a^{1/2}}\int\cot^{n-1}u\ \mathrm{d}u$$ I then used the $\cot^nu$ reduction formula to find $$I_n=\frac{-b^{1/2-n}}{a^{1/2}}\bigg(\frac{\cot^{n-2}u}{n-2}+\int\cot^{n-3}u\ \mathrm{d}u\bigg)$$ $$I_n=\frac{-b^{1/2-n}\cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$ Which is a reduction formula, but not the reduction formula.

Could someone provide a derivation of the reduction formula? Thanks.

  • $\begingroup$ I think you've found the reduction formula depends on $b$. $\endgroup$ – Nosrati Nov 15 '18 at 4:48
  • $\begingroup$ @Nosrati how so? $\endgroup$ – clathratus Nov 15 '18 at 4:49
  • $\begingroup$ Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b \geq 0$. $\endgroup$ – Travis Willse Nov 15 '18 at 20:03

Hint The appearance of the term in $\frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get $$I_m = u v - \int v \,du = \frac{x}{(a x^2 + b)^m} + 2 m \int \frac{a x^2 \,dx}{(a x^2 + b)^{m + 1}} .$$ Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.

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    $\begingroup$ Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1) $\endgroup$ – clathratus Nov 15 '18 at 5:01
  • $\begingroup$ I'm glad you found it useful, cheers! $\endgroup$ – Travis Willse Nov 15 '18 at 5:05
  • $\begingroup$ Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick? $\endgroup$ – clathratus Nov 15 '18 at 18:52
  • $\begingroup$ I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating! $\endgroup$ – Travis Willse Nov 15 '18 at 19:40
  • $\begingroup$ Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick. $\endgroup$ – clathratus Nov 15 '18 at 19:41

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