I'd like to solve the integral
$$I := \int_{0}^{\pi/2} \ln(\sec^{2}(x) + \tan^{4}(x)) \mathrm{d}x$$
using the method of differentiating under the integral sign. So, first substituting $\tan x=u$, we get $\sec^2x \mathrm{d}x = \mathrm{d}u$, meaning $\mathrm{d}x = \frac{\mathrm{d}u}{\sec^2(\tan^{-1} u)} = \frac{\mathrm{d}u}{1+u^2}$. So, we have
$$I = \int_{0}^{\infty} \ln(1+u^{2} + u^{4}) \frac{\mathrm{d}u}{1+u^2}.$$
Let's define $I(t)$ as
$$I(t) := \int_{0}^{\infty} \ln(1+t[u^{2} + u^{4}]) \frac{\mathrm{d}u}{1+u^2},$$
where $I(t=0) = 0$. Then, differentiating w.r.t. $t$ yields
$$I'(t) = \int_{0}^{\infty} \frac{[u^{2} + u^{4}]\mathrm{d}u}{(1+t[u^{2} + u^{4}])(1+u^2)} = \int_{0}^{\infty} \frac{u^{2}\mathrm{d}u}{1+tu^{2}(1 + u^{2})}.$$
Let $x = u^2$. Then, $\mathrm{d}u = \frac{\mathrm{d}x}{2\sqrt{x}}$, leading to
$$I'(t) = \frac{1}{2t}\int_{0}^{\infty} \frac{\sqrt{x}\mathrm{d}x}{(x+\frac{1}{2})^2 + (\frac{1}{t}-\frac{1}{4})}.$$
Does this integral even converge? If so, how can one evaluate it? If otherwise, what have I done wrong here?
Best Answer
You can continue from $$I'(t) = \int_{0}^{\infty} \frac{u^{2}}{1+tu^{2}(1 + u^{2})} \,du$$ $$\frac{u^{2}}{1+tu^{2}(1 + u^{2})}=\frac 1 t \frac{u^2}{(u^2-a)(u^2-b)}$$ where $$a=\frac{-t-\sqrt{(t-4)t} }{2 t} \qquad \text{and} \qquad b=\frac{-t+\sqrt{(t-4)t} }{2 t}$$ Using partial fraction decomposition $$I'(t)=\frac 1{t(a-b)}\int_0^\infty \left(\frac{a}{u^2-a}-\frac{b}{u^2-b} \right)\,du$$ which are simple. Using the bounds
$$I'(t)=\frac{\pi \left(\frac{2}{\sqrt{t-\sqrt{(t-4)t} }}-\frac{2}{\sqrt{t+\sqrt{(t-4)t} }}\right)}{2 \sqrt{2} \sqrt{(t-4)t} }$$ Integrating again $$I(t)=\pi \left(\tanh ^{-1}\left(\frac{\sqrt{t-\sqrt{(t-4)t} }}{\sqrt{2}}\right)+\tanh ^{-1}\left(\frac{\sqrt{t+\sqrt{(t-4)t} }}{\sqrt{2}}\right)\right)$$
Since $I(0)=0$, the result is then $$I=\pi \left(\tanh ^{-1}\left(\sqrt{\frac{1}{2} \left(1-i \sqrt{3}\right)}\right)+\tanh ^{-1}\left(\sqrt{\frac{1}{2} \left(1+i \sqrt{3}\right)}\right)\right)$$ that is to say $$\color{blue}{\large I=\pi \log \left(2+\sqrt{3}\right)}$$