1

$$I_{\alpha}=\int_{0}^{\pi/2}\mathrm dx {\sin(2x)\over 1+\tan^{\alpha}x}={1\over 2}\tag1$$

$\alpha \ge 0$

The indefinite integral for $\alpha=1,2,3...$

$$I_1={1\over 4}\sin(2x)-{1\over 4}\cos(2x)-{1\over 2}\log[\sin x+\cos x]+C$$

$$I_2=-{1\over 2}\cos^4(x)+C$$

$$I_3=-{1\over 4}\sin(2x)+{1\over 4}\cos(2x)-{2\over 3}\log[2-\cos 2x]-{1\over 6}\ln[\sin x +\cos x]+C$$

The indefinite integral looks complicate for $\alpha$, how can $(1)$ gives ${1\over 2}?$

  • If you plot the functions for various $\alpha$, you get what looks like a probability distribution. Then you can consider twice the integrand to be a probability distribution with a parameter $P(x;\alpha) = \frac{2\sin(2x)}{1+\tan^{\alpha}x}$, which is normalised (it's integral equals $1$) on the support of $0$ to $\pi/2$. $\tan(x)$ is monotonically increasing in this interval, and $\sin(x)$ is bounded. – Benedict W. J. Irwin Apr 16 '18 at 13:15

3 Answers3

3

Hint:

As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$

if $\displaystyle I=\int_a^bf(x)dx$

$$I+I=\int_a^b\{f(x)dx+f(a+b-x)\}dx=?$$

lab bhattacharjee
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  • See also: https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Apr 16 '18 at 13:06
  • @SUHANA, What is $$f(a+b-x)=?$$ here? – lab bhattacharjee Apr 16 '18 at 13:13
  • I don't know if this is useful but there is a change of variables that comes in mind:

    as $\sin(2x)=2t/(1+t^2)$ where $t=\tan(x)$ you have: $$I_\alpha=\int_0^{\pi/2} \frac{2 \tan(x)}{(1+\tan(x)^\alpha)(1+\tan(x)^2)} dx$$ with $t=\tan(x)$ you obtain: $$I_\alpha=\int_0^\infty \frac{2t dt}{(1+t^2)^2(1+t^\alpha)}$$ and $u=t^2$: $$I_\alpha=\int_0^\infty \frac{1}{(1+u)^2 (1+u^\frac{\alpha}{2})} du$$

     – Delta-u Apr 16 '18 at 13:17
  • @SUHANA, $$\int_0^{\pi/2}\sin2x\left(\dfrac1{1+\tan^mx}+\dfrac1{1+\cot^mx}\right)$$ Now $$\dfrac1{1+\tan^mx}+\dfrac1{1+\cot^mx}=?$$ – lab bhattacharjee Apr 16 '18 at 13:28
1

By differentiating $I_\alpha$ with respect to $\alpha$, one obtains: $$I’=\int^{\pi /2}_0 \sin(2x)\frac{\ln(\tan(x))\tan^\alpha (x)}{(1+\tan^{\alpha}(x))^2}dx$$

By the substitution $x \to \pi/2-x$, one obtains: $$I’=\int^{\pi /2}_0 \sin(2x)\frac{\ln(\cot(x))\cot^\alpha (x)}{(1+\cot^{\alpha}(x))^2}dx=-\int^{\pi /2}_0 \sin(2x)\frac{\ln(\tan(x))\tan^\alpha (x)}{(1+\tan^{\alpha}(x))^2}dx=-I’$$

Since $I’=-I’$, $I’=0$.

Thus, no matter what value of $\alpha$ you substitute in, the integral is still the same.

Substitute in $\alpha=0$, you will obtain $I=\frac12$ easily.

What is even more interesting is the case $\alpha =\infty$. For $x<\pi/4$, $\tan(x)<0$, $\tan^\infty (x)=0$, thus the integrand reduces to $\sin(2x)$; for $x>\pi/4$, $\tan^\infty (x)=\infty$, the integrand is zero. So, $$I_\infty=\int^{\pi/4}_0 \sin(2x)dx=\frac12$$ Similarly, $$I_{-\infty}=\int^{\pi/2}_{\pi/4}\sin(2x)dx=\frac12$$

Szeto
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0

As @labbhattacharjee suggested, use the property

$\int_{a}^{b}f(x)=\int_{a}^{b}f(a+b-x)$

$\implies I=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\tan^\alpha x}dx=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\cot^\alpha x}dx$

$\implies 2I=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\tan^\alpha x}+\frac{\sin 2x}{1+\cot^\alpha x}dx$

$\implies 2I= \int_{0}^{\pi/2}(\frac{\sin 2x(1+\tan^\alpha x)}{1+\tan^\alpha x}+\frac{\sin 2x(1+\cot^\alpha x)}{1+\cot^\alpha x})dx$

$\implies 2I=\int_{0}^{\pi/2}(\frac{\sin 2x(1+\tan^\alpha x+\cot^\alpha x +1)}{1+\tan^\alpha x+\cot^\alpha x +1})dx $

$\implies 2I =\int_{0}^{\pi/2}(\sin 2x) dx $

$\therefore 2I=1$

$\therefore I=\frac{1}{2}$

SUHAN A
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