Evaluate the improper integral $\int_{-\infty}^{+\infty} \frac{\ln(1+x^{4})}{1+x^{2}} dx$

Question

How do I evaluate this integral ? I am a student who just studied calculus 1 and recently definite integral.

$$ \int_{-\infty}^{+\infty} \frac{\ln(1+x^{4})}{1+x^{2}} dx $$

I substituted both $x = \tan t$ and $x = \frac1{t}$ but could not proceed further.

Answer 1

A polynomial of the form $1+x^{2n}$ can always be written as the product of $n$ quadratic factors of the form $1-2x\cos\theta + x^2$. So that suggests we consider the integral $$ I(\theta) = \int_{-\infty}^\infty \frac{\ln(1-2x\cos\theta+x^2)}{1+x^2}dx. $$ Now, if we differentiate this with respect to $\theta$, we'll get a rational function of $x$, and the integral can then be evaluated through partial fractions: \begin{eqnarray} I'(\theta) &=& \int_{-\infty}^\infty \frac{2x\sin\theta}{(1-2x\cos\theta+x^2)(1+x^2)}dx \\&=&\tan\theta\int_{-\infty}^\infty\left(\frac{1}{1-2x\cos\theta+x^2}- \frac{1}{1+x^2}\right)dx \\ &=& \pi\frac{1-\sin\theta}{\cos\theta}=\pi\frac{\cos\theta}{1+\sin\theta} = \pi\frac{d}{d\theta}\ln(1+\sin\theta). \end{eqnarray} This gives us the $\theta$ dependence of $I$, but we need the value of $I$ at a particular $\theta$ to set the constant of integtration. We can evaluate the case $\theta = \pi/2$ with clever use of trigonometric identities and variable substitutions: \begin{eqnarray} I\left(\frac{\pi}{2}\right)& = &\int_{-\infty}^\infty \frac{\ln(1+x^2)}{1+x^2} = -4\int_{0}^{\pi/2}\ln(\cos\phi)d\phi = -4\int_{0}^{\pi/2}\ln(\sin\phi)d\phi \\ &=& -2\int_0^{\pi}\ln(\sin\phi)d\phi =-4\int_0^{\pi/2}\ln(\sin 2\psi)d\psi = -4\int_0^{\pi/2}\ln(2\sin\psi\cos\psi)d\psi \\ &=& -2\pi\ln 2-4\int_{0}^{\pi/2}\ln(\cos\psi)d\psi-4\int_{0}^{\pi/2}\ln(\sin\psi)d\psi \\ &=& -2\pi\ln 2 + 2I\left(\frac{\pi}{2}\right) \end{eqnarray} Thus $I(\pi/2) = 2\pi \ln2$, and we have $$ I(\theta) = \pi\ln[2(1+\sin\theta)]. $$ From here it's easy to show $$ \int_{-\infty}^\infty \frac{\ln(1+x^4)}{1+x^2}dx = I\left(\frac{\pi}{4}\right) + I\left(\frac{3\pi}{4}\right) = 2\pi\ln(2+\sqrt{2}) $$

Answer 2

Factorize $1+x^4= (1+i x^2)(1-ix^2)$ and then utilize the integral $$\int_{-\infty}^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= 2\pi\ln(1+a^{\frac12}) $$ to obtain

\begin{align} \int_{-\infty}^{\infty}\frac{\ln (1+x^{4})}{1+x^2}dx &=\int_{-\infty}^{\infty}\frac{\ln (1+i x^{2})}{1+x^{2}}+\frac{\ln (1-i x^{2})}{1+x^{2}}\ dx\\ &=2\pi \ln\left[(1+i^{\frac12})(1+(-i)^{\frac12})\right]=2\pi\ln (2+\sqrt2) \end{align}

Answer 3

Hint: Under $x=\tan t$, you have \begin{eqnarray} &&\int_{-\infty}^{+\infty} \frac{\ln(1+x^{4})}{1+x^{2}} dx\\ &=&\int_{-\frac\pi2}^{\frac\pi2}\ln(1+\tan^4t)dt\\ &=&\int_{-\frac\pi2}^{\frac\pi2}\ln(\cos^4+\sin^4t)dt-4\int_{-\frac\pi2}^{\frac\pi2}\ln(\cos t)dt\\ &=&\int_{-\frac\pi2}^{\frac\pi2}\ln\bigg(\frac34+\frac14\cos(4t)\bigg)dt-4\int_{-\frac\pi2}^{\frac\pi2}\ln(\cos t)dt\\ &=&\int_{-\frac\pi2}^{\frac\pi2}\ln\bigg(3+\cos(4t)\bigg)dt+2\pi\ln2\\ &=&2\int_{-\pi}^{\pi}\ln(3+\cos t)dt+2\pi\ln2\\ \end{eqnarray} Here $$\int_{-\frac\pi2}^{\frac\pi2}\ln(\cos t)dt=-\pi\ln2 $$ is used. For the first term, you can define $$ I(a)=\int_{-\pi}^{\pi}\ln(a+\cos t)dt, a\ge 1 $$ and then use Feyman's trick to get it. I omit the detail. (Evaluating $\int_{0}^{\pi}\ln (1+\cos x)\, dx$, Evaluate $\int_0^{{\pi}/{2}} \log(1+\cos x)\, dx$)

Answer 4

I don't see how to evaluate this integral using the techniques from a typical first integral calculus course.

We'll use two standard techniques: Differentiating under the integral sign, and applying the Residue Theorem to an appropriate contour.

Consider the family $I(a)$ of integrals defined by the $$I(a) = \int_{-\infty}^\infty \frac{\log[1 + (a x)^4]\,dx}{1 + x^2}$$ Our integral is $I(1)$.

Differentiating $I(a)$ with respect to $a$ and rearranging gives $$I'(a) = 4 a^3 \int_{-\infty}^\infty \frac{x^4 \,dx}{(1 + a x^4) (1 + x^2)} = \frac{4 a^3}{1 + a^4} \left[\int_{-\infty}^\infty \frac{dx}{1 + x^2} + \int_{-\infty}^{\infty} \frac{x^2 - 1}{1 + a^4 x^4}\,dx\right]$$

The first integral in brackets is $\arctan x \vert_{-\infty}^\infty = \frac\pi2$. To evaluate the second integral, we could use the method of partial fractions to decompose the integral and the Fundamental Theorem of Calculus, but the antiderivative is messy. So we instead apply the Residue Theorem to a standard semicircular contour $\Gamma_R$. The residues of $f(z) := \frac{z^2 - 1}{1 + a^4 z^4}$ inside $\Gamma_R$ (for large enough $R$) are at $\exp \frac{\pi i}4, \exp \frac{3\pi i}4$: \begin{align} \operatorname{Res}\left(f(z), \frac{\exp\frac{\pi i}{4}}{a}\right) &= \frac{1}{2 \sqrt 2 a^3} \left[(1 + a^2) - (1 - a^2) i\right] \\ \operatorname{Res}\left(f(z), \frac{\exp\frac{3\pi i}{4}}{a}\right) &= \frac{1}{2 \sqrt 2 a^3} \left[-(1 + a^2) - (1 - a^2) i\right] \end{align} So, $$\int_{-\infty}^\infty \frac{x^2 - 1}{1 + a^4 x^4} \,dx = 2 \pi i \sum_{z_i} \operatorname{Res}\left(f(z), z_i\right) = \frac{\pi (1 - a^2)}{\sqrt 2 a^3}.$$

Assembling the above ingredients gives $$I'(a) = 2 \pi \frac{\sqrt 2 + 2 a}{1 + \sqrt 2 a + a^2} ,$$ so our integral has value \begin{multline}\require{cancel} I(1) = I(1) - \cancelto{0}{I(0)} = \int_0^1 I'(a) \,da = 2 \pi \int_0^1 \frac{\sqrt 2 + 2 a}{1 + \sqrt 2 a + a^2} \,da \\ = 2 \pi \log(1 + \sqrt 2 a + a^2)\vert_0^1 = \boxed{2 \pi \log(2 + \sqrt 2)} .\end{multline}

Answer 5

Using the identity $\ln \left(x^2+y^2\right)=2 \Re(\ln (x+y i))$ to reduce the power $4$ of $x$ in the numerator makes the life easier. $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x = & \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x \\ = & \Re\left(\int_{-\infty}^{\infty} \frac{\ln \left(x^2+i\right)}{1+x^2}\right)\\=&\Re \int_{-\infty}^{\infty} \frac{\ln \left(x^2+\left(\frac{1+i}{\sqrt{2}}\right)^2\right)}{1+x^2} \end{aligned} $$ Using the particular case $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+a^2\right)}{1+x^2} d x=2 \pi \ln (1+a)$ in the post, we have $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x & =\Re\left[2 \pi \ln \left(1+\frac{1+i}{\sqrt{2}}\right)\right] \\ & =2 \pi \Re\left[\ln \left(1+\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]\\ & =\pi \ln \left(1+\frac{1}{2}+\sqrt{2}+\frac{1}{2}\right) \\ & =\pi \ln (2+\sqrt{2}) \end{aligned} $$

Answer 6

Trigonometric Reduction $$ \begin{align} &\int_0^\infty\frac{\log\left(1+x^4\right)}{1+x^2}\,\mathrm{d}x\\ &=\int_0^{\pi/2}\log\left(1+\tan^4(x)\right)\,\mathrm{d}x\tag{1a}\\ &=\int_0^{\pi/2}\log\left(\cos^4(x)+\sin^4(x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1b}\\ &=\int_0^{\pi/2}\log\left(1-2\sin^2(x)\cos^2(x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1c}\\ &=\int_0^{\pi/2}\log\left(1-\frac12\sin^2(2x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1d}\\ &=\frac12\int_0^\pi\log\left(1-\frac12\sin^2(x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1e}\\ &=\int_0^{\pi/2}\log\left(1-\frac12\sin^2(x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1f}\\ &=\int_0^{\pi/2}\log\left(1-\frac12\cos^2(x)\right)\,\mathrm{d}x-4\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{1g}\\ \end{align} $$ Explanation:
$\text{(1a):}$ substitute $x\mapsto\tan(x)$
$\text{(1b):}$ add and subtract the integral of $\log\left(\cos^4(x)\right)$
$\text{(1c):}$ $\cos^4(x)+\sin^4(x)=\left(\cos^2(x)+\sin^2(x)\right)^2-2\sin^2(x)\cos^2(x)$
$\text{(1d):}$ $\sin(2x)=2\sin(x)\cos(x)$
$\text{(1e):}$ substitute $x\mapsto x/2$ in the left integral
$\text{(1f):}$ apply the symmetry of $\sin(x)$ about $\frac\pi2$
$\text{(1g):}$ substitute $x\mapsto\frac\pi2-x$ in the left integral

---

Computation of the Left Integral $$ \begin{align} &\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x\tag{2a}\\ &=\int_0^{\pi/2}\cos^{2n-1}(x)\,\mathrm{d}\sin(x)\tag{2b}\\ &=(2n-1)\int_0^{\pi/2}\sin^2(x)\cos^{2n-2}(x)\,\mathrm{d}x\tag{2c}\\ &=(2n-1)\int_0^{\pi/2}\cos^{2n-2}(x)\,\mathrm{d}x-(2n-1)\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x\tag{2d}\\ &=\frac{2n-1}{2n}\int_0^{\pi/2}\cos^{2n-2}(x)\,\mathrm{d}x\tag{2e} \end{align} $$ Explanation:
$\text{(2b):}$ prepare to integrate by parts
$\text{(2c):}$ integrate by parts
$\text{(2d):}$ $\sin^2(x)=1-\cos^2(x)$
$\text{(2e):}$ add $\frac{2n-1}{2n}$ times $\text{(2a)}$ to $\frac1{2n}$ times $\text{(2c)}$

Apply $(2)$ inductively to obtain $$ \int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x=\frac{\pi/2}{4^n}\binom{2n}{n}\tag3 $$ To apply $(3)$ to the integral, we will use the series $$ \begin{align} \sum_{n=0}^\infty\binom{2n}{n}x^n&=\frac1{\sqrt{1-4x}}\tag{4a}\\ \sum_{n=1}^\infty\binom{2n}{n}x^{n-1}&=\frac4{\sqrt{1-4x}\left(1+\sqrt{1-4x}\right)}\tag{4b}\\ \sum_{n=1}^\infty\frac1n\binom{2n}{n}x^n&=2\log\left(\frac2{1+\sqrt{1-4x}}\right)\tag{4c}\\ \end{align} $$ Explanation:
$\text{(4a):}$ the Extended Binomial Theorem
$\text{(4b):}$ subtract $1$ and divide by $x$
$\text{(4c):}$ integrate using $\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{1-4x}=-\frac2{\sqrt{1-4x}}$

Finally, $$ \begin{align} \int_0^{\pi/2}\log\left(1-\alpha\cos^2(x)\right)\,\mathrm{d}x &=-\sum_{n=1}^\infty\frac{\alpha^n}n\int_0^{\pi/2}\cos^{2n}(x)\,\mathrm{d}x\tag{5a}\\ &=-\sum_{n=1}^\infty\frac{\alpha^n}n\frac{\pi/2}{4^n}\binom{2n}{n}\tag{5b}\\ &=\pi\log\left(\frac{1+\sqrt{1-\alpha}}2\right)\tag{5c} \end{align} $$ Explanation:
$\text{(5a):}$ apply the series for $\log(1-x)$
$\text{(5b):}$ apply $(3)$
$\text{(5c):}$ apply $\text{(4c)}$ with $x=\frac\alpha4$

---

Computation of the Right Integral

Although we can use $(5)$, with $\alpha=1$, to compute the following integral, the following approach not only displays a different method, but also is more direct and simpler. $$ \begin{align} \int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x &=\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{6a}\\ &=\frac12\int_0^{\pi/2}\log\left(\frac12\sin(2x)\right)\,\mathrm{d}x\tag{6b}\\ &=\frac12\int_0^{\pi/2}\log\left(\frac12\sin(x)\right)\,\mathrm{d}x\tag{6c}\\[3pt] &=-\frac\pi2\log(2)\tag{6d} \end{align} $$ Explanation:
$\text{(6a):}$ substitute $x\mapsto\frac\pi2-x$
$\text{(6b):}$ average the two sides of $\text{(6a)}$
$\text{(6c):}$ substitute $x\mapsto x/2$
$\phantom{\text{(6c):}}$ apply the symmetry of $\sin(x)$ about $\frac\pi2$
$\text{(6d):}$ subtract $\text{(6a)}$ from two times $\text{(6c)}$

---

The Complete Integral $$ \begin{align} &\int_{-\infty}^\infty\frac{\log\left(1+x^4\right)}{1+x^2}\,\mathrm{d}x\\ &=2\int_0^\infty\frac{\log\left(1+x^4\right)}{1+x^2}\,\mathrm{d}x\tag{7a}\\ &=2\int_0^{\pi/2}\log\left(1-\frac12\cos^2(x)\right)\,\mathrm{d}x-8\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{7b}\\ &=2\pi\log\left(\frac{2+\sqrt2}4\right)-8\left(-\frac\pi2\log(2)\right)\tag{7c}\\[3pt] &=2\pi\log\left(2+\sqrt2\right)\tag{7d} \end{align} $$ Explanation:
$\text{(7a):}$ the integrand is even
$\text{(7b):}$ apply $\text{(1g)}$
$\text{(7c):}$ apply $(5)$ with $\alpha=\frac12$ and $(6)$
$\text{(7d):}$ simplify

Answer 7

$\frac{\ln(x^4+1)}{x^2+1}=\sum_{n=1}^4\frac{\ln(x-w^n)}{x^2+1}$ where $w=e^{\frac{\pi i}{4}}$. Applying the residue theorem with the closed lower semi-circle contour for $n=1,2$ and with the upper one for $n=3,4$, the answer will be $$2\pi i\left(-\sum_{n=1}^2\frac{\ln(-i-w^n)}{-2i}+ \sum_{n=3}^4\frac{\ln(i-w^n)}{2i}\right)= \pi\left(\ln(2+\sqrt2)+\ln(2+\sqrt2)\right)\\=2\pi\ln(2+\sqrt2).$$