1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $n - 1$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.