If $\displaystyle f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz,$ where $x>-1$. Then value of $\displaystyle f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right) = $

$\bf{My\; Try::}$ Given $\displaystyle f(x) = \int_{0}^{\frac{\pi}{4}}\ln \left(1+x\cdot \tan z\right)dz$

$\displaystyle \Rightarrow f^{'}(x) = \int_{0}^{\frac{\pi}{4}}\frac{\tan z}{1+x\cdot \tan z}dz = \int_{0}^{\frac{\pi}{4}}\frac{\frac{2\tan \frac{z}{2}}{1-\tan^2 \frac{z}{2}}}{1+x\cdot \frac{2\tan \frac{z}{2}}{1-\tan^2 \frac{z}{2}}}dz = \int_{0}^{\frac{\pi}{4}}\frac{2\tan \frac{z}{2}}{1-\tan^2\frac{z}{2}+2x\cdot \tan \frac{z}{2}}dz$

Now Let $\displaystyle \tan \frac{z}{2}=t$, Then $\displaystyle dz=\frac{2}{1+t^2}dt$

So $\displaystyle f^{'}(x) = \int_{0}^{\frac{\pi}{4}}\frac{4t}{1-t^2+2x\cdot t}dt$

Now How can I solve after that

Please Help me


  • $\begingroup$ You need to change the bounds of the integral after the substitution for $t$. Also, you might prefer to rewrite $\frac{\tan z}{1 + x \tan z}$ by polynomial division before integrating. Finally, the last integral will need to be solved by integration by partial fractions. $\endgroup$ – user119191 Jan 5 '14 at 13:29
  • $\begingroup$ Thanks user119191 for you hint $\endgroup$ – juantheron Jan 7 '14 at 14:50

There is a somewhat simpler way to procede: once you get to $$f^\prime(x) = \int_0^{\pi/4} \frac{\tan z}{1+x \tan z} dz,$$ make the substitution $u = 2 z,$ so the integral becomes $$\frac12\int_0^{\pi/2} \frac{\tan u/2}{1+x \tan u/2} du.$$ Now make the substitution $\tan u/2 = t,$ to get $$\int_0^1 \frac{t}{(1+ x t)(1+t^2)} dt.$$

This is easily integrated by partial fractions, to give you $$ \frac{\pi x-4 \log (x+1)+\log (4)}{4 x^2+4}. $$ You can go on from there, though I would not describe this as super fun.


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