Can this be correct $I_{\alpha}=\int_{0}^{\pi/2}\mathrm dx {\sin(2x)\over 1+\tan^{\alpha}x}={1\over 2}?$

Question

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

$\alpha \ge 0$

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 ${1\over 2}?$

Answer 1

Hint:

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

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

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

Answer 2

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, $$I_{-\infty}=\int^{\pi/2}_{\pi/4}\sin(2x)dx=\frac12$$

Answer 3

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 $

$\therefore 2I=1$

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