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I would like to evaluate (using elementary methods if possible) : (for $a>0,\ b>0$)

$$ I_n=\int_0^{\pi/2} \frac{1}{( a\cos^2x+b\sin^2x)^n} \, dx,\quad \ n=1,2,3,\ldots $$ I thought about using $u=\tan(x)$ or $u=\frac{\pi}{2}-x$ but did not work. wolfram alpha evaluates the indefinite integral but not definite integral???

Quanto
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4 Answers4

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Hint:Use Feynman’s Trick: differentiate the integral with respect to the parameters $a$ and $b$, and it can be shown that:

$$\frac{\partial {{I}_{n}}}{\partial a}+\frac{\partial {{I}_{n}}}{\partial b}=-n{{I}_{n+1}}$$ This recursion can be re-written alternatively as: $${{I}_{n}}=-\frac{1}{n-1}\left( \frac{\partial {{I}_{n-1}}}{\partial a}+\frac{\partial {{I}_{n-1}}}{\partial b} \right),\quad n=2,3,...$$ and notice that ${{I}_{1}}$ can be evaluated rather easily using $u=\tan \left( x \right)$ to get ${{I}_{1}}=\frac{\pi }{2\sqrt{ab}}$.

  • Is it possible to use Feynman's Trick here $$\int_0^\frac{\pi}{2} \frac{\cos(A \cos(t))}{a^2 \sin^2 t + b^2 \cos^2 t}dt?$$ – Dinesh Shankar Aug 16 '18 at 20:36
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    I am not sure, let me try –  Aug 18 '18 at 07:12
  • Thank you. If you can, let me know, so I'll post this as a new question. – Dinesh Shankar Aug 18 '18 at 11:53
  • assume $I\left( A \right)=\int_{0}^{\pi /2;}{\frac{\cos \left( A\cos \left( t \right) \right)}{{{a}^{2}}{{\sin }^{2}}\left( t \right)+{{b}^{2}}{{\cos }^{2}}\left( t \right)}dt}$ –  Aug 18 '18 at 12:24
  • differentiate w.r.s to $A$ n times$${{I}^{\left( n \right)}}\left( A \right)=\int_{0}^{{\pi }/{2};}{\frac{{{\cos }^{n}}\left( t \right)\cos \left( {n\pi }/{2};+A\cos \left( t \right) \right)}{{{a}^{2}}{{\sin }^{2}}\left( t \right)+{{b}^{2}}{{\cos }^{2}}\left( t \right)}dt}$$ –  Aug 18 '18 at 12:25
  • Now \begin{align} & {{I}^{\left( n \right)}}\left( 0 \right)={{\left( -1 \right)}^{\frac{n}{2}}}\int_{0}^{\pi /2;}{\frac{{{\cos }^{n}}\left( t \right)}{{{a}^{2}}{{\sin }^{2}}\left( t \right)+{{b}^{2}}{{\cos }^{2}}\left( t \right)}dt,\quad if\quad n=0,2,4,...} \ & {{I}^{\left( n \right)}}\left( 0 \right)=0,\quad if\quad n=1,3,5,... \ \end{align}$ $ –  Aug 18 '18 at 12:30
  • if we assume that $I\left( A \right)$ has a Maclaurin series representation,then $$I\left( A \right)=\sum\nolimits_{m=0}^{\infty }{\frac{{{I}^{\left( 2m \right)}}\left( 0 \right)}{\left( 2m \right)!}{{A}^{2m}}}$$ where $${{I}^{\left( 2m \right)}}\left( 0 \right)={{\left( -1 \right)}^{m+1}}\frac{\sqrt{\pi },\Gamma \left( \frac{1}{2}+m \right){}{2}{{F}{1}}(\frac{1}{2},1,1+m,1-\frac{{{a}^{2}}}{{{b}^{2}}})}{2{{b}^{2}}m!}$$ –  Aug 18 '18 at 12:41
  • this is complicated,,,,, sorry ,,,,,, i tried my best –  Aug 18 '18 at 12:44
  • besides i am not sure if it is valid to assume that $I\left( A \right)$ has a Maclaurin series? any way if you are going to post a question about this, write this attempt, i would like to see other answers or corrections. –  Aug 18 '18 at 12:50
  • This is a nice answer. But unfortunately, this does not work for me. Apparently, this integral was solved in terms of elementary functions by an acquaintance of mine. I do not know how he did it. There is another version of this integral. Are you interested? – Dinesh Shankar Aug 18 '18 at 12:53
  • yes i am, tell me –  Aug 18 '18 at 12:54
  • Here the integral: $$ \int_{0}^{\pi/2}\frac{1-a \cos^2x}{A \cos^2x+B\sin^2x} \mathrm e^{-\frac{a}{2}\cos^2x}. $$ I am sure that this can be solved in terms of elementary functions. – Dinesh Shankar Aug 18 '18 at 13:01
  • i agree and i will take my time –  Aug 18 '18 at 13:02
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    I will post a question about the first integral. –  Aug 18 '18 at 14:36
  • https://math.stackexchange.com/questions/2886810/evaluation-of-int-0-pi-2-frac-cos-left-a-cos-t-righta2 –  Aug 18 '18 at 15:15
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    If you are interested. Here is a discussion about this second integral. – Dinesh Shankar Aug 18 '18 at 19:54
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Per geometric summation \begin{align} \sum_{n\ge 1}I_n t^n=& \int_0^{\frac{\pi}{2}} \sum_{n\ge 1}\frac{t^n}{(a \cos^2x+ b \sin^2x)^n}dx\\ =&\int_0^{\frac{\pi}{2}}\frac{t}{a\cos^2x+ b\sin^2x-t}dx =\frac{\pi}{2}\frac{t}{\sqrt{(a-t )(b-t )}}\\ =&\sum_{n\ge 1}\frac{\pi}{2^{2n-1}}\sum_{i+j+1=n} \frac{1}{a^{i+1/2} b^{j+1/2}} {2i\choose i}{2j\choose j} t^n \end{align} where $ \frac{1}{\sqrt{a-t}}=\sum_{k\ge 0}\frac{1}{2^{2k} a^{k+1/2}} {2k\choose k }t^k$ is applied in the last step. Thus $$I_n= \frac{\pi}{2^{2n-1}\sqrt{ab}}\sum_{i+j+1=n} \frac{{2i\choose i}{2j\choose j}}{a^{i} b^{j}} $$

Quanto
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You may just use the residue theorem. $a$ and $b$ are interchangeable via $x\mapsto\frac{\pi}{2}-x$, so it is safe to assume $c\stackrel{\text{def}}{=}\frac{b}{a}\in(0,1)$ (since the case $a=b$ is trivial) and study $$ I_n(a,b) = \frac{1}{a^n}\int_{0}^{\pi/2}\frac{d\theta}{(\cos^2\theta+c\sin^2\theta)^n}\stackrel{\theta\mapsto\arctan u}{=}\frac{1}{a^n}\int_{0}^{+\infty}\frac{du}{(1+u^2)\left(\frac{1+cu^2}{1+u^2}\right)^n}$$ which is $$ \frac{1}{2a^n}\int_{\mathbb{R}}\frac{(1+u^2)^{n-1}}{(1+cu^2)^{n}}\,du =\frac{\pi i}{a^n}\operatorname*{Res}_{u=i/\sqrt{c}}\frac{(1+u^2)^{n-1}}{(1+cu^2)^n}.$$ $u=\frac{i}{\sqrt{c}}$ is clearly a pole of order $n$ for $\frac{(1+u^2)^{n-1}}{(1+cu^2)^n}$, hence the RHS equals $$\frac{\pi i}{(n-1)! a^n c^n}\lim_{u\to \frac{i}{\sqrt{c}}} \frac{d^{n-1}}{du^{n-1}}\frac{(1+u^2)^{n-1}}{\left(u+\frac{i}{\sqrt{c}}\right)^n}. $$

Jack D'Aurizio
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This is not an answer but it is too long for a comment.

For the antiderivative, Wolfram Alpha (and other CAS) return, for $$J_n(x)=(n-1)(a-b) I_n(x)$$s the messy expression $$2^{n-1} \csc (2 x) \sqrt{\frac{(a-b) \sin ^2(x)}{a}} \sqrt{\frac{(b-a) \cos ^2(x)}{b}}$$ $$ ((a-b) \cos (2 x)+a+b)^{1-n}$$ $$ F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b+(a-b) \cos (2 x)}{2 b},\frac{a+b+(a-b) \cos (2 x)}{2 a}\right)$$ where appears

The problem is that both $J_n\left(\frac{\pi }{2}\right)$ and $J_n\left(0\right)$ result in indeterminate forms and that the limits need to be worked.

These would be $$J_n\left(\frac{\pi }{2}\right)=-\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}}\, b^{1-n}\, \Gamma (2-n)}{2 \, \Gamma \left(\frac{3}{2}-n\right)}\, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{b}{a}\right)$$ $$J_n\left(0\right)=\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}} \,a^{1-n}\, \Gamma (2-n)}{2\, \Gamma \left(\frac{3}{2}-n\right)}\, \, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{a}{b}\right)$$

Have fun !

Claude Leibovici
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