 deiteresfp

2021-12-30

Evaluate the integral
${\int }_{0}^{\pi /2}\frac{1}{1+{\mathrm{tan}}^{\alpha }x}dx$ MoxboasteBots5h

Expert

The integral can be solved by using the fact that $\mathrm{tan}\left(\frac{\pi }{2}-x\right)=\frac{1}{\mathrm{tan}x}$
${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{b}f\left(a+b-x\right)dx$
Let
$I\left(\alpha \right)={\int }_{0}^{\frac{\pi }{2}}\frac{dx}{1+{\mathrm{tan}}^{\alpha }x}$
then$I\left(\alpha \right)={\int }_{0}^{\frac{\pi }{2}}\frac{dx}{1+{\mathrm{tan}}^{\alpha }\left(\frac{\pi }{2}+0-x\right)}$
$={\int }_{0}^{\frac{\pi }{2}}\frac{dx}{1+\frac{1}{{\mathrm{tan}}^{\alpha }x}}$
$={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{tan}}^{\alpha }x}{1+{\mathrm{tan}}^{\alpha }x}dx$
Adding the two $I{\left(\alpha \right)}^{\prime }s$ yields
$2I\left(\alpha \right)={\int }_{0}^{\frac{\pi }{2}}\frac{1}{1+{\mathrm{tan}}^{\alpha }x}dx+{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{tan}}^{\alpha }x}{1+{\mathrm{tan}}^{\alpha }x}dx$
$={\int }_{0}^{\frac{\pi }{2}}dx$
$=\frac{\pi }{2}$
$I\left(\alpha \right)=\frac{\pi }{4}$ Mason Hall

Expert

Set $I\left(\alpha \right)\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\frac{\pi }{2}}f\left(\alpha ,x\right)dx$ where $f\left(\alpha ,x\right):\mathbb{R}×\left(0,\frac{\pi }{2}\right)\to \mathbb{R}$ is defined as $f\left(\alpha ,x\right)\phantom{\rule{0.222em}{0ex}}=\frac{1}{1+{\mathrm{tan}}^{\alpha }\left(x\right)}$
Since both $\frac{d}{d\alpha }f\left(\alpha ,x\right)=-\frac{{\mathrm{tan}}^{\alpha }\left(x\right)\mathrm{log}\left(\mathrm{tan}\left(x\right)\right)}{{\left(1+{\mathrm{tan}}^{\alpha }\left(x\right)\right)}^{2}}$ and $f\left(\alpha ,x\right)$ are continuous on $\left(0,\frac{\pi }{2}\right)\to \mathbb{R}$ we can apply the Leibniz integral rule obtaining
${I}^{\prime }\left(\alpha \right)=\frac{d}{d\alpha }{\int }_{0}^{\frac{\pi }{2}}f\left(\alpha ,x\right)dx={\int }_{0}^{\frac{\pi }{2}}\frac{d}{d\alpha }f\left(\alpha ,x\right)dx=0$
since $\frac{d}{d\alpha }f\left(\alpha ,\frac{\pi }{4}+x\right)$$=\frac{d}{d\alpha }f\left(\alpha ,\frac{\pi }{4}-x\right)$ for $x\in \left[0,\frac{\pi }{4}\right)$.
Jence $I\left(\alpha \right)$ is constant on $\mathbb{R}$ and
$I\left(\alpha \right)=I\left(0\right)={\int }_{0}^{\frac{\pi }{2}}\frac{dx}{1+1}=\frac{\pi }{4}$ Vasquez

Expert

Well, this problem has a nice symmetry. The integral can be rewritten as,
$I={\int }_{0}^{\pi /2}\frac{{\mathrm{cos}}^{\alpha }\left(x\right)}{{\mathrm{sin}}^{\alpha }\left(x\right)+{\mathrm{cos}}^{\alpha }\left(x\right)}dx$
By the property of definite integrals this Integral is same as,
${\int }_{0}^{\pi /2}\frac{{\mathrm{cos}}^{\alpha }\left(\frac{\pi }{2}-x\right)}{{\mathrm{sin}}^{\alpha }\left(\frac{\pi }{2}-x\right)+{\mathrm{cos}}^{\alpha }\left(\frac{\pi }{2}-x\right)}dx$
${\int }_{0}^{\pi /2}\frac{{\mathrm{sin}}^{\alpha }\left(x\right)}{{\mathrm{sin}}^{\alpha }\left(x\right)+{\mathrm{cos}}^{\alpha }\left(x\right)}$
$2I={\int }_{0}^{\pi /2}dx$
$2I=\frac{\pi }{2}$
$I=\frac{\pi }{4}$

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