Monday Math 159

Find I=\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\ln\tan{x}}dx

First, we note that this is an improper integral, as the integrand is not only undefined for the limits of integration, but for the value x=π/4 as well. So, we consider the limit of the integrand as it approaches these values.

First, as x approaches zero (from above) the numerator \ln\left(\frac{\tan^{\pi/2}x+1}{2}\right) approaches \ln\left(\frac{0+1}{2}\right)=-\ln{2}, while the denominator approaches -∞. Therefore, the integrand goes to zero at the lower limit of integration.

For x=π/4, both \ln\tan{x} and \ln\left(\frac{\tan^{\pi/2}x+1}{2}\right) equal zero, and for x=π/2, both approach +∞. Thus, in both cases we have indeterminate forms amenable to L’Hôpital’s rule. And since
\begin{array}{rcl}\frac{\frac{d}{dx}\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\frac{d}{dx}\ln\tan{x}}&=&\frac{\frac{\frac{d}{dx}\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\left(\frac{\tan^{\pi/2}x+1}{2}\right)}}{\frac{\frac{d}{dx}\tan{x}}{\tan{x}}}\\&=&\frac{\frac{\frac{\pi}{2}\tan^{\frac{\pi}{2}-1}{x}\sec^2{x}}{\tan^{\pi/2}x+1}}{\frac{\sec^2{x}}{\tan{x}}}\\&=&\frac{\pi}{2}\frac{\tan^{\pi/2}x}{\tan^{\pi/2}x+1}\\&=&\frac{\pi}{2}\left[1-\frac{1}{\tan^{\pi/2}x+1}\right]\end{array}

So
\lim_{x\to\pi/4}\frac{\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\ln\tan{x}}=\lim_{x\to\pi/4}\frac{\pi}{2}\left[1-\frac{1}{\tan^{\pi/2}x+1}\right]=\frac{\pi}{2}\left[1-\frac{1}{1+1}\right]=\frac{\pi}{4}

and the singularity at x=π/4 is removable, and the (improper) integral is unaffected; and
\lim_{x\to\pi/2-}\frac{\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\ln\tan{x}}=\lim_{x\to\pi/2-}\frac{\pi}{2}\left[1-\frac{1}{\tan^{\pi/2}x+1}\right]=\frac{\pi}{2}\left[1-0\right]=\frac{\pi}{2}

And the integrand has a finite limit at the upper limit of integration as well. Thus, despite the undefined values, the integrand is sufficiently well-behaved that we need not be much concerned with the improper nature of the integral.

There does not appear to be a clear antiderivative for the integrand, and attempting a substitution like u=\tan{x} doesn’t appear to help. What we can do, though is an indirect method, which begins with the substitution u=\frac{\pi}{2}-x. Then du=-dx and \tan{x}=\tan\left(\frac{\pi}2-u\right)=\cot{u}=\frac1{\tan{u}}. Thus,
\begin{array}{rcl}I&=&\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}x+1}{2}\right)}{\ln\tan{x}}dx\\&=&\int_{\pi/2}^{0}\frac{\ln\left(\frac{\frac1{\tan^{\pi/2}u}+1}{2}\right)}{\ln\left(\frac1{\tan{u}}\right)}(-du)\\&=&\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}u+1}{2\tan^{\pi/2}u}\right)}{-\ln\tan{u}}du\\&=&\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}u+1}{2}\right)-\ln\left(\tan^{\pi/2}u\right)}{-\ln\tan{u}}du\\&=&\int_0^{\pi/2}\frac{\frac{\pi}{2}\ln\tan{u}-\ln\left(\frac{\tan^{\pi/2}u+1}{2}\right)}{\ln\tan{u}}du\\&=&\int_0^{\pi/2}\frac{\pi}{2}du-\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}u+1}{2}\right)}{\ln\tan{u}}du\\I&=&\frac{\pi^2}{4}-\int_0^{\pi/2}\frac{\ln\left(\frac{\tan^{\pi/2}u+1}{2}\right)}{\ln\tan{u}}du\end{array}
But this last integral is just our original I, only with u as the variable of integration instead of x. Therefore,
\begin{array}{rcl}I&=&\frac{\pi^2}{4}-I\\2I&=&\frac{\pi^2}{4}\\I&=&\frac{\pi^2}{8}\end{array}

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