A question about $\int_{0}^{\pi/2} \ln(\sec^{2}(x) + \tan^{4}(x)) \mathrm{d}x$

Question

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?

Answer 1

Alternatively, differentiate $$I(a)= \int_{-\pi/2}^{\pi/2}\ln(\sec^2x+2\sin a\tan x)dx, \>\>\>\>\> I’(a)= -\frac{\pi\sin a}{1+\cos a} $$ to integrate \begin{align} & \int_{0}^{\pi/2} \ln(\sec^{2}x + \tan^{4}x) {dx}\\ =& \ \frac12\int_{-\pi/2}^{\pi/2}\ln[(\sec^2x+\tan x)(\sec^2x-\tan x)]dx\\ =& \int_{-\pi/2}^{\pi/2}\ln(\sec^2x+\tan x)dx=I(\frac\pi6) =I(0) +\int_0^{\pi/6}I’(a)da\\ =& \int_{-\pi/2}^{\pi/2}\ln(\sec^2x)dx -\int_0^{\pi/6} \frac{\pi\sin a}{1+\cos a} da=\pi \ln(2+\sqrt3) \end{align}

Answer 2

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 $$\color{blue}{\large I=\pi \log \left(2+\sqrt{3}\right)}$$

Answer 3

Continuing from your derivative step, instead by symmetry we have

$$I'[t] = \frac{1}{2}\int_{-\infty}^\infty \frac{u^2\:du}{1+tu^2+tu^4} = \frac{1}{2}\int_{-\infty}^\infty \frac{du}{t+2\sqrt{t}+t\left(u-\frac{1}{u\sqrt{t}}\right)^2}$$

We can now use Glasser's Master theorem to state that $\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)dx = \int_{-\infty}^\infty f(x)dx$ for positive $a$

$$I'[t] = \frac{1}{2t}\int_{-\infty}^\infty\frac{du}{\left(1+\frac{2}{\sqrt{t}}\right)+u^2} = \frac{\pi}{2t\sqrt{1+\frac{2}{\sqrt{t}}}}$$

Integrating, this suggests using the substitution $s = \sqrt{1+\frac{2}{\sqrt{t}}}$, which gives

$$I[t] = \int \frac{2\pi \:ds}{1-s^2} = 2\pi\coth^{-1}s+C = 2\pi\coth^{-1}\sqrt{1+\frac{2}{\sqrt{t}}}+C$$

being cognizant of using the appropriate branch between $\tanh^{-1}$ and $\coth^{-1}$. To find $C$, note that

$$I[0] = 0 = 2\pi\coth^{-1}(\infty)+C = 0 + C$$

$$\implies \boxed{I[t] = 2\pi\coth^{-1}\sqrt{1+\frac{2}{\sqrt{t}}}}$$

And the integral we want is $I[1]$

$$I[1] = \boxed{2\pi\coth^{-1}\sqrt{3}}$$

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EDIT: Just for fun, we can simplify the $I[t]$ formula using hyperbolic identities

$$I[t] = 2\pi\coth^{-1}\sqrt{1+\frac{2}{\sqrt{t}}} = 2\pi\sinh^{-1}\sqrt[4]{\frac{t}{4}} = \pi \cosh^{-1}\left(1+\sqrt{t}\right)$$

from $\coth^2\tau - 1 = \frac{1}{\sinh^2\tau}$ first and then $\cosh2\tau = 1+2\sinh^2\tau$ second.

Answer 4

Set $$I(t) := \int_0^\infty \frac{\log(t^4 u^4 + t^2 u^2 + 1)}{u^2 + 1} \,du,$$ so that our desired integral is $I(1)$. Differentiating under the integral sign gives $$\require{cancel}I'(t) = \int_0^\infty \frac{4 t^3 u^4 + 2 t u^2}{(t^4 u^4 + t^2 u^2 + 1) (u^2 + 1)} du = \frac{\pi (2 t + \sqrt 3)}{t^2 + \sqrt 3 t + 1}.$$ The quartic factor in the denominator of the integrand factors into quadratics as $$(t^2 u^2 + t u + 1)(t^2 u^2 - t u + 1),$$ so the above evaluation of the previous integral is a straightforward (if somewhat tedious) application of the Method of Partial Fractions. The computation can be significantly reduced by exploiting that the integrand is even.

In any case, integrating again gives \begin{align} I(1) &= \cancelto{0}{I(0)} + \int_0^1 I'(t) \,dt \\ &= \pi \int_0^1 \frac{2 t + \sqrt 3}{t^2 + \sqrt 3 t + 1} \,dt \\ &= \pi \log (t^2 + \sqrt{3} t + 1)\vert_0^1 \\ &= \pi \log (2 + \sqrt{3}) \\ &= \pi \operatorname{arcosh} 2. \end{align}

Answer 5

In fact \begin{eqnarray} &&\sec^{2}(x) + \tan^{4}(x)=\frac{\cos^{2}(x)+\sin^4(x)}{\cos^{4}(x)}\\ &=&\frac{\cos^2x+(1-\cos^2(x))^2}{\cos^{4}(x)}=\frac{1-\cos^2x+\cos^4(x)}{\cos^{4}(x)}\\ &=&\frac{1-\frac{1+\cos(2x)}2+\frac{(1+\cos(2x))^2}4}{\cos^{4}(x)}=\frac{3+\cos^2(2x)}{4\cos^{4}(x)}\\ &=&\frac{3+\frac{1+\cos(4x)}{2}}{4\cos^{4}(x)}=\frac{7+\cos(4x)}{8\cos^{4}(x)}\\ \end{eqnarray} and hence \begin{eqnarray} I &=& \int_{0}^{\pi/2} \ln(\sec^{2}(x) + \tan^{4}(x)) \mathrm{d}x\\ &=&\int_{0}^{\pi/2} \ln(7+\cos(4x)) \mathrm{d}x-\int_{0}^{\pi/2} \ln(8)\mathrm{d}x-4\int_{0}^{\pi/2}\ln(\cos(x))\mathrm{d}x\\ &=&\frac14\int_{0}^{2\pi} \ln(7+\cos(x)) \mathrm{d}x-\frac{3\pi}2 \ln(2)-4\int_{0}^{\pi/2}\ln(\cos(x))\mathrm{d}x\\ \end{eqnarray} Now using $$ \int_{0}^{\pi/2}\ln(\cos(x))\mathrm{d}x=-\frac\pi2\ln(2), $$ and $$ \int_{0}^{\pi}\ln(1+a\cos(x))\mathrm{d}x=\frac\pi2\ln(1+\sqrt{1-a^2})$$ from How to Integrate $\int_0^\pi \ln(1+\alpha \cos(x)) \,\mathrm{d}x$, it is easy to get the answer which is $\pi\ln(2+\sqrt3)$.

Answer 6

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \ln\left(\sec^2x + \tan^4x\right) \, dx \\ &= \frac12 \int_0^\infty \frac{\ln\left(x^2+x+1\right)}{\sqrt x (x+1)} \, dx \tag1 \\ &= i \pi \underset{z=-1}{\operatorname{Res}} \frac{\ln\left(z^2+z+1\right)}{\sqrt z (z+1)} + \frac{\sqrt3\,\pi}2 \int_1^\infty \frac{x-1}{x^2-x+1} \, \frac{dx}{\sqrt x} \tag2 \\ &= \sqrt3 \, \pi \int_0^1 \frac{1-x^2}{1-x^2+x^4} \, dx \tag3 \\ &= - \sqrt3 \, \pi \int_0^1 \frac{1 - \frac1{x^2}}{\left(x+\frac1x\right)^2 - 3} \, dx \tag4 \\ &= \sqrt3\,\pi \int_2^\infty \frac{dy}{y^2 - 3} = \boxed{\pi \log\left(2+\sqrt3\right)} \tag5 \end{align*}$$

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• $(1)$ substitute $x\mapsto\arctan \sqrt x$
• $(2)$ retrace the steps of this answer and use the residue theorem
• $(3)$ substitute $x\mapsto \dfrac1{x^2}$
• $(4)$ divide through by $\dfrac1{x^2}$ and complete the square
• $(5)$ substitute $y=x+\dfrac1x$

Answer 7

Define $I(b) = \int_0^{\pi/2} \log(b \sec^2 \theta + \tan^4 \theta) d\theta$.

Then, $$I'(b) = \int_0^{\pi/2} \frac{\sec^2 \theta}{b \sec^2 \theta + \tan^4 \theta} d\theta$$

$$ = \int_0^{\infty} \frac{1}{b (x^2+1) + x^4} d x$$

$$ = b^{-3/4}\int_0^{\infty} \frac{1}{1 + \sqrt{b} y^2 + y^4} d y$$

$$ =\frac{b^{-3/4}}{2}\int_0^{\infty} \frac{(1+\frac{1}{y^2})-(1-\frac{1}{y^2})}{\sqrt{b} + y^2 + \frac{1}{y^2}} d y$$

$$ =\frac{b^{-3/4}}{2}\int_0^{\infty} \frac{(1+\frac{1}{y^2})}{2 + \sqrt{b} + (y- \frac{1}{y})^2} d y - \frac{b^{-3/4}}{2}\int_0^{\infty} \frac{(1-\frac{1}{y^2})}{(y+ \frac{1}{y})^2 - (2-\sqrt{b})} d y$$

$$ =b^{-3/4} \left(\frac{\pi}{2 \sqrt{2+\sqrt{b}}} - 0\right)$$

Integrating with respect to $b$,

$$I(b) = \int \frac{\pi}{2} \frac{b^{-\frac{3}{4}}}{\sqrt{2 + \sqrt{b}}} db + c$$ $$ = 2 \pi \log\left(\frac{b^{1/4}+\sqrt{2+\sqrt{b}}}{\sqrt{2}}\right)+ c$$

Since $I(0) = 4 \int_0^{\pi/2} \log(\tan \theta) d\theta = 0$, we must have $c=0$.

Therefore, $\int_0^{\pi/2} \log( \sec^2 \theta + \tan^4 \theta) d\theta$

$$=I(1) = 2\pi \log(\frac{\sqrt{3}+1}{\sqrt{2}})$$ $$ = \pi \log(2+\sqrt{3})$$