Determining Suitable Contour for $\int_0^{\infty} \dfrac{\log (x) }{x^2 +1 } dx$

[Amer]

what is the suitable contour for the integral
$\displaystyle \int_0^{\infty} \dfrac{\log (x) }{x^2 +1 } dx $

and why ? or what is the method to determine a good one?

 

[ThePerfectHacker]

what is the suitable contour for the integral
$\displaystyle \int_0^{\infty} \dfrac{\log (x) }{x^2 +1 } dx $

and why ? or what is the method to determine a good one?

Let $\varepsilon > 0$ be small and $R>0$ be large.
1) Start at $i\varepsilon$, in the clockwise direction draw semicircle $C$ centered at $0$ that ends at $-i\varepsilon$.
2) Draw rays emanating at $\pm i\varepsilon$, endpoints of $C$, to the left, so $-t\pm i\varepsilon $ for $t\geq 0$.
3) Draw circle $\Gamma$ centered at $0$ of radius $R$.
4) The rays intersect $\Gamma$ at points $\alpha,\beta$ with $\text{Im}(\alpha) > \text{Im}(\beta)$.

Now define $\Gamma_1$ as contour starting at $\beta$, travel counterclockwise along $\Gamma$ until reaching $\alpha$. Travel linearly along ray until reaching $i\varepsilon$. Then move clockwise along $C$ until reaching $-i\varepsilon$. Then travel linearly along ray until returning back to $\alpha$.

Define the function,
$$ f(z) = \frac{\log z}{z^2 + 1} $$
Where $\log z$ is the principal logarithm and integrate,
$$ \oint_{\Gamma_1} f(z) ~ dz $$

 

[chisigma]

what is the suitable contour for the integral
$\displaystyle \int_0^{\infty} \dfrac{\log (x) }{x^2 +1 } dx $

and why ? or what is the method to determine a good one?

In...

http://mathhelpboards.com/calculus-10/improper-integral-involving-ln-6103-post27786.html#post27786

... in elementary fashion it has been demonstrated that...

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{1 + x^{2}}\ dx = 0\ (1)$

... but it is also been demonstrated that no suitable contour of integration in the complex plane for the integral (1) exists... Kind regards $\chi$ $\sigma$

 

[ThePerfectHacker]

The best way of computing that integral is to rather use a substitution $x\mapsto 1/x$.

Let $I = \int_0^{\infty} \frac{\log x}{x^2+1} ~ dx$. Now, let $y = \frac{1}{x}$, so that $y' = -\frac{1}{x^2}$ and $x = \frac{1}{y}$. Rewrite,
$$ \frac{\log x}{x^2+1} = \frac{\log x}{x^2+1} \cdot \left( -\frac{1}{x^2} \right) \cdot \left( -x^2 \right) = \frac{ \log \frac{1}{y}}{ \frac{1}{y^2} + 1 } \cdot y' \cdot \left( -\frac{1}{y^2} \right) = \frac{\log y}{y^2+1} \cdot y' $$
Futhermore, $y(0) = \infty$ and $y(\infty) = 0$, so that,
$$ I = \int_0^{\infty} \frac{\log x}{x^2+1} ~ dx = \int_{\infty}^{0} \frac{\log y}{y^2+1} ~ dy = - I $$
This implies that $I=0$.

 

[polygamma]

Let $\varepsilon > 0$ be small and $R>0$ be large.
1) Start at $i\varepsilon$, in the clockwise direction draw semicircle $C$ centered at $0$ that ends at $-i\varepsilon$.
2) Draw rays emanating at $\pm i\varepsilon$, endpoints of $C$, to the left, so $-t\pm i\varepsilon $ for $t\geq 0$.
3) Draw circle $\Gamma$ centered at $0$ of radius $R$.
4) The rays intersect $\Gamma$ at points $\alpha,\beta$ with $\text{Im}(\alpha) > \text{Im}(\beta)$.

Now define $\Gamma_1$ as contour starting at $\beta$, travel counterclockwise along $\Gamma$ until reaching $\alpha$. Travel linearly along ray until reaching $i\varepsilon$. Then move clockwise along $C$ until reaching $-i\varepsilon$. Then travel linearly along ray until returning back to $\alpha$.

Define the function,
$$ f(z) = \frac{\log z}{z^2 + 1} $$
Where $\log z$ is the principal logarithm and integrate,
$$ \oint_{\Gamma_1} f(z) ~ dz $$

That combination of function and contour won't work. The log term will vanish.

You either need to consider $ \displaystyle f(z) = \frac{\log^{2} z}{1+z^{2}} $, or you could move the branch cut from the negative real axis to the negative imaginary axis and integrate around a contour that consists of the real axis (indented at the origin) and the upper half of the circle $|z|=R$.

 

[alyafey22]

If we integrate the function

$$f=\frac{\log^2(z)}{z^2+1}$$

along the contour

View attachment 2411

$$\left(\oint_{C_R}+\oint_{c_r} \right)f(z)\,dz + \int^R_r \frac{\log^2(x)}{x^2+1}\,dx-\int^{R}_r\frac{(\log(x)+2\pi i)^2}{x^2+1}\,dx =2\pi i \sum_{i=1}^2\text{Res}(f,z_i)$$

Taking $R\to \infty$ and $r \to 0$

Then we have

$$-4\pi i\int^\infty_0 \frac{\log(x)}{x^2+1}\,dx+4\pi^2\int^{\infty}_0\frac{1}{x^2+1}\,dx =2\pi i \sum_{i=1}^2\text{Res}\left(\frac{\log^2(z)}{z^2+1},\pm i \right)$$

$$\sum_{i=1}^2\text{Res}\left(\frac{\log^2(z)}{z^2+1},\pm i \right)=\left(\frac{\log^2(i)}{2i} -\frac{\log(-i)^2}{2i}\right)=-\frac{\pi^2}{8i}+\frac{9\pi^2}{8i}=-i\pi^2$$

$$-4\pi i\int^\infty_0 \frac{\log(x)}{x^2+1}\,dx+4\pi^2\int^{\infty}_0\frac{1}{x^2+1}\,dx =2\pi^3$$

Hence we have that

$$\int^{\infty}_0\frac{1}{x^2+1}\,dx=\frac{\pi}{2}$$

$$\int^\infty_0 \frac{\log(x)}{x^2+1}\,dx=0$$