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A logarithm integral

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Find the following integral :

\(\displaystyle \int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx\)
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

TheBigBadBen

Active member
May 12, 2013
84
Find the following integral :

\(\displaystyle \int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx\)
I am sure that there's some nice little trick to all this, but here I go anyway...

u-substition:
$$
u = \ln(x);\; du = dx/x \rightarrow\\
\int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx=
\int^{\infty}_{-\infty} \frac{u}{e^{2u}+a^2}\,e^u du\\
=\int^{\infty}_{-\infty} \frac{u}{e^{u}+a^2e^{-u}}\, du
$$

Well, that's my stab at it for now. More on that when I'm less sleepy.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Find the following integral :

\(\displaystyle \int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx\)
Setting $\frac{x}{a}= t$ the integral becomes...

$$I= \frac{\ln a}{a}\ \int_{0}^{\infty} \frac{d t}{1 + t^{2}}\ dt + \frac{1}{a} \int_{0}^{\infty} \frac{\ln t}{1+ t^{2}}\ dt = \frac{\pi}{2}\ \frac{\ln a}{a} + \frac{1}{a} (\int_{0}^{1} \frac{\ln t}{1+ t^{2}}\ dt + \int_{1}^{\infty} \frac{\ln t}{1+ t^{2}}\ dt)\ (1)$$

... and setting in the last integral in (1) $\xi= \frac{1}{t}$ we obtain...

$$I= \frac{\pi}{2}\ \frac{\ln a}{a} + \frac{1}{a}\ (\int_{0}^{1} \frac{\ln t}{1+ t^{2}}\ dt - \int_{0}^{1} \frac{\ln \xi}{1+ \xi^{2}}\ d \xi) = \frac{\pi}{2}\ \frac{\ln a}{a}\ (2)$$

Kind regards

$\chi$ $\sigma$
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle F(b) = \int^{\infty}_0 \frac{x^b}{x^2+a^2}\, \, \) ----(1)

\(\displaystyle F(b) = \frac{1}{a^2}\,\int^{\infty}_0 \frac{x^b}{\frac{x^2}{a^2}+1} \, dx\)

Let \(\displaystyle \frac{x^2}{a^2} = t\)

\(\displaystyle F(b) = \frac{a^b}{2|a|} \,\int^{\infty}_0 \frac{t^{\frac{b-1}{2}}}{t+1} \, dt = \frac{a^b \, \pi }{2|a| \sin\left( \pi \frac{b+1}{2}\right)}\)

\(\displaystyle F(b)=\frac{a^b \, \pi }{2|a| \sin\left( \pi \frac{b+1}{2}\right)}\)

Differentiate wrt to $b$

\(\displaystyle F'(b) = \frac{a^b \, \pi \, \ln|a| }{2|a| \sin\left( \pi \frac{b+1}{2}\right)} - \frac{a^b \, \pi^2 }{4|a| } \cot^2 \left(\pi \frac{b+1}{2}\right)\)

\(\displaystyle F'(0) = \frac{ \, \pi \, \ln|a| }{2|a|} \)

Differentiating (1) wrt to \(\displaystyle b\) we get :

\(\displaystyle F(b) = \int^{\infty}_0 \frac{x^b \ln(x) }{x^2+a^2} \, dx\)

\(\displaystyle F'(0) = \int^{\infty}_0 \frac{\ln(x) }{x^2+a^2} \, dx\, \, \) ----(2)

By (1) , (2) we get what we want \(\displaystyle \square\) .
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
There is a complex analysis approach , if someone is interested . I will try to post it later .
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is a complex approach

\(\displaystyle f(z) = \frac{\log_0^2(z)}{z^2+a^2}\)

Where the logarithm has a branch cut on the positive real axis and the integration is along a circle with radius \(\displaystyle R\).

We will divide the contour into two parts because the function is not analytic in the whole contour , so we can't apply the residue theory directly. We can easily prove that the sum of the integration along the two parts will eventually be \(\displaystyle 2\pi i \, \text{Res}(f(z))\) . The sum of integration along the two tilted lines will be equal to $0$.

contour2.PNG

The picture shows two separated parts while there should be no distance between the two contours . So the two lines should be on the x-axis but I moved them a little bit for easy illustration and to identify direction .



Integration along the x-axis

\(\displaystyle \int^R_0 \frac{\ln^2(x)}{x^2+a^2}\, dx\)


Integration along the x-axis ( opposite direction )

\(\displaystyle -\int^R_0 \frac{(\ln(x)+2\pi i)^2}{x^2+a^2}\, dx=-\int^R_0 \frac{\ln^2(x)+4\pi i \ln(x)-4\pi^2}{x^2+a^2}\, dx\)


Integration along the Circle

\(\displaystyle \int_{C_R}\frac{\log_0 ^2(z)}{z^2+a^2}\, dz \leq 2 \pi R \frac{\ln^2|R|+ 2\pi \ln|R| +4\pi^2}{R^2-a^2} \to 0 \,\,\,, R \to \infty\)


Sum of Residues


Here we are assuming that \(\displaystyle a>0\) to easily find the residues


\(\displaystyle \text{Res}(f(z) ; ai) +\text{Res}(f(z) ; -ai) =\frac{\log_0^2(ai)}{2ai}-\frac{\log_0^2(-ai)}{2ai} \)


\(\displaystyle \frac{\left(\ln(a)+\frac{\pi}{2}i \right)^2}{2ai}-\frac{\left(\ln(a)+\frac{3\pi}{2}i \right)^2}{2ai}= \frac{-2 \pi \ln(a)i +2\pi^2}{2ai}\)


Now take \(\displaystyle R\to \infty\) and sum the contours

\(\displaystyle -4\pi\text{P.V} \int^{\infty}_0 \frac{ \ln(x)}{x^2+a^2} \, dx -4\pi^2 \text{P.V} \int^{\infty}_0 \frac{dx}{x^2+a^2}\, dx = 2\pi i\left( \frac{-2 \pi \ln(a)i +2\pi^2}{2ai}\right)\)


Comparing the imaginary parts


\(\displaystyle -4\pi\text{P.V} \int^{\infty}_0 \frac{ \ln(x)}{x^2+a^2} \, dx= -\frac{2 \pi ^2 \ln (a)}{a}\)


\(\displaystyle \text{P.V}\int^{\infty}_0 \frac{ \ln(x)}{x^2+a^2}\, dx = \frac{\pi \ln (a)}{2a} \,\,\, a>0\)


We also get by comparing the real part


\(\displaystyle 4\pi^2 \text{P.V} \int^{\infty}_0 \frac{dx}{x^2+a^2} = \frac{2\pi^3}{a}\)

\(\displaystyle \int^{\infty}_0 \frac{dx}{x^2+a^2} = \frac{\pi}{2a}\)
 
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