Calculate $\int_0^\infty \frac{\ln(t)}{1+t^3}dt$

Question

How can I evaluate $$\int_0^\infty \frac{\ln(t)}{1+t^3}dt$$ I've seen similar posts with $1+t^2$ instead of $1+t^3$. But I'm not sure whether they could help me in this case. I'm thinking of using the residue theorem. But I'm stuck on correctly applying it.

Answer 1

\begin{align} &\int_0^\infty\frac{\ln t}{t^3+1}{d}t\\ =&\int_0^1\frac{(1-t)\ln t}{t^3+1}{d}t =\int_0^1\frac{(t^2-t+1)\ln t}{t^3+1}{d}t - \int_0^1{\frac{t^2\ln t}{t^3+1}}\overset{t^3\to t}{dt}\\ =&\int_0^1\frac{\ln t}{t+1}{d}t-\frac19 \int_0^1\frac{\ln t}{t+1}{d}t =\frac89\cdot \left(-\frac{\pi^2}{12}\right) =-\frac{2\pi^2}{27} \end{align}

where $\int_0^1\frac{\ln t}{t+1}{d}t=-\frac{\pi^2}{12}$

Answer 2

Let's compute a more general form.

$$I(a,n,m)=\int_0^\infty \frac{t^a\ln^n t}{1+t^m}dt$$

Let

$$F(a)=\int_0^\infty \frac{t^a}{1+t^m}dt\Longrightarrow I(a,n,m)=\frac{d^n}{da^n}F(a)$$

Let $u=t^m$

$$F(a)=\frac 1m\int_0^\infty \frac{u^{\frac{1+a}{m}-1}}{1+u}du=\frac1mB\left(\frac{1+a}{m},1-\frac{1+a}{m}\right)=\frac{\pi}{m\cdot\sin\left( \frac{\pi(1+a)}{m}\right)}$$ hence

$$\boxed{\int_0^\infty \frac{t^a\ln^n t}{1+t^m}dt=\frac{d^n}{da^n}\left(\frac{\pi}{m\cdot\sin\left( \frac{\pi(1+a)}{m}\right)}\right)}$$

For you case, $a=0, n=1, m=3$

$$\int_0^\infty \frac{\ln t}{1+t^3}dt=\frac{d}{da}\left(\frac{\pi}{3\cdot\sin\left( \frac{\pi(1+a)}{3}\right)}\right)_{a=0}=-\frac{2\pi^2}{27}$$

Answer 3

\begin{align} I &=\int_0^{1} \frac{\log t}{1+t^3}dt+\int_1^{\infty} \frac{\log t}{1+t^3}dt\\ & =\int_0^{1} \frac{\log t}{1+t^3}dt-\int_0^1 \frac{t\log t}{1+t^3}dt \text{, $t\to1/t$ in the second}\\ &=\int_0^{1} \frac{(1-t)\log t}{1+t^3}dt\\ &=\sum_{j=0}^{\infty}\int_0^1(-t^3)^j(1-t)\log t dt\\ &=\sum_{j=0}^{\infty}\frac{-(-1)^j}{(3j+1)^2}-\sum_{j=0}^{\infty}\frac{-(-1)^j}{(3j+2)^2}=\cdots=-\frac{2\pi^2}{27}\\ \end{align} The rest of fun for you; you should trust the number $-\frac{2\pi^2}{27}$ provided by @Quanto - I did :-)

Answer 4

As a usual trick, $$\begin{align*} \int_{0}^{\infty} \frac{\log t}{1+t^3}\, dt &= \int_{0}^{1} \frac{\log t}{1+t^3}\, dt + \underbrace{\int_{1}^{\infty} \frac{\log t}{1+t^3}\, dt}_{t\rightarrow \frac{1}{t}} \\ &= \int_{0}^{1} \frac{(1-t)\log t}{1+t^3}\, dt \\ &= \int_{0}^{1} (1-t)\log t \sum_{n=0}^{\infty} (-t)^{3n}\, dt \\ &= \sum_{n=0}^{\infty} (-1)^n \underbrace{\int_{0}^{1} (t^{3n}-t^{3n+1})\log t \, dt}_{u=\log t, \ dv= (t^{3n}-t^{3n+1})\, dt\\ du= \frac{1}{t}\, dt, \ v = \frac{t^{3n+1}}{3n+1}-\frac{t^{3n+2}}{3n+2}}\\ &=\sum_{n=0}^{\infty} (-1)^n \left(\log t \left(\frac{t^{3n+1}}{3n+1}-\frac{t^{3n+2}}{3n+2}\right)\Bigg\rvert_{0}^{1}- \int_{0}^{1} \left(\frac{t^{3n}}{3n+1}-\frac{t^{3n+1}}{3n+2}\right)\, dt\right) \\ &= -\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2}\right).\end{align*}$$ Now,

The alternating sum of reciprocal of squares has the value, $$\frac{1}{1^2}- \frac{1}{2^{2}}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots = \frac{\pi^2}{12}.$$

$\textit{Proof:}$ We have $$\begin{align*}\frac{1}{1^2}- \frac{1}{2^{2}}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots &= \left(\frac{1}{1^2}+\frac{1}{2^{2}}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots \right)-2\times \frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^{2}}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots \right)\\ &= \frac{\pi^2}{6}-\frac{\pi^2}{12}\\ &= \frac{\pi^2}{12}. \end{align*} $$ Note that the alternating sum of reciprocal of squares can be rewritten as $$\sum_{n=0}^{\infty} (-1)^n \left(\frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2}+\frac{1}{(3n+3)^2}\right)= \frac{\pi^2}{12}.$$ Therefore, $$\begin{align*} \int_{0}^{\infty} \frac{\log t }{1+t^3} \, dt &= -\frac{\pi^2}{12} + \frac{1}{9}\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}\\ &= -\frac{\pi^2}{12}+\frac{\pi^2}{9\cdot 12} \\ &= - \frac{2\pi^2}{27}. \quad \blacksquare\end{align*}$$

Answer 5

Here is an outline of the contour integral method to find

$$I = \int_0^\infty \frac{\log p \, dp}{p^3+1}$$

Choose a contour that goes counterclockwise with these "pieces":

$$\begin{aligned} \gamma_a&: \text{ along the x-axis from }\epsilon \text{ to } R \\ \gamma_R&: \text{ a third of a circle of radius }R\\ \gamma_b&: \text{ the line from }Re^{2\pi I /3} \text{ to } \epsilon e^{2\pi I/3}\\ \gamma_\epsilon &:\text{ a third of a circle of radius }\epsilon \end{aligned}$$ $$ C= \gamma_a \cup \gamma_R \cup \gamma_b \cup \gamma_\epsilon$$ Now consider $$\oint_C \frac{\log z \, dz}{z^3+1}= 2\pi i \text{ Res}_{z=e^{i \pi / 3}} \left( \frac{\log z}{z^3+1}\right)$$

$$\oint_C = \int_{\gamma_a}+\int_{\gamma_R}+\int_{\gamma_b}+\int_{\gamma_\epsilon}$$

The integrals along the arcs go to zero as $R\to \infty$and $\epsilon \to 0$. We are left with

$$\int_0^\infty \frac{\log p\, dp}{p^3+1}- e^{2\pi i/3}\int_0^\infty \frac{ \log p + 2\pi I/3}{p^3+1} \, dp = 2\pi i \text{ Res}_{z=e^{i \pi / 3}} \left( \frac{\log z}{z^3+1}\right)$$

$$(1-e^{2\pi i/3}) I - e^{2\pi i/3} \int_0^\infty \frac{dp}{p^3+1}= 2\pi i \text{ Res}_{z=e^{i \pi / 3}} \left( \frac{\log z}{z^3+1}\right)$$

Let $$\begin{aligned} \alpha &= \int_0^\infty \frac{dp}{p^3+1} = \frac{2\pi}{3\sqrt{3}}\\ \beta &=\text{ Res}_{z=e^{i \pi / 3}} \left( \frac{\log z}{z^3+1}\right)= -\frac{\pi}{9}e^{5\pi i/6} \end{aligned} $$

Combining terms:

$$I = \frac{e^{2\pi i/3} \alpha + 2\pi i \beta}{1-e^{2\pi i/3}} = - \frac{2\pi^2}{27}.$$

Answer 6

Slightly different approach using complex analysis: taking a branch cut along the positive real axis, we integrate

$$\oint_C \frac{\log^{\color{red}2}z}{1+z^3} \, dz = \oint_C \frac{\left(\log |z| + i\arg z\right)^2}{1+z^3}\,dz$$

over an indented circular contour $C$ composed of

• a small circle of radius $\varepsilon$ around $z=0$,
• a large circle or radius $R\gg1$ around $z=0$, and
• two opposing line segments $L$ and $L'$ running parallel and to either side of the positive real axis that join the arcs.

It's easy to show the integrals over the first two components vanish as $\varepsilon\to0^+$ and $R\to\infty$, respectively. Above the cut, as the line segment gets closer to the real axis, we have

$$\int_L \frac{\log^2z}{1+z^3} \, dz \to \int_0^\infty \frac{\log^2x}{1+x^3} \, dx$$

and below the cut,

$$\int_{L'} \frac{\log^2z}{1+z^3} \, dz \to -\int_0^\infty \frac{\left(\log x + i2\pi\right)^2}{1+x^3} \, dx$$

so that together, they contribute

$$\int_{L\cup L'} \frac{\log^2z}{1+z^3} \, dz \to 4\pi\int_0^\infty \frac{\pi - i \log x}{1+x^3} \, dx$$

and we see the integral we want in the imaginary part. Apply the residue theorem. There are three poles at the third roots of $-1$, $z\in\left\{e^{i\pi/3},e^{i\pi},e^{i5\pi/3}\right\}$. We find that

$$i 2\pi \cdot \frac{4\pi^2\left(1-i\,3\sqrt3\right)}{27} = 4\pi\int_0^\infty \frac{\pi - i \log x}{1+x^3} \, dx \\ \implies \int_0^\infty \frac{\log x}{1+x^3} \, dx=\boxed{-\frac{2\pi^2}{27}}$$