Integral of Logarithm: Find & Solve

Finally, we get the desired result as a combination of the two integrals:\int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx = \text{P.V} \int^{\infty}_0 \frac{ \ln(x)}{x^2+a^2}\, dx + \int^{\infty}_0 \frac{dx}{x^2+a^2} = \frac{\pi \ln (a)}{2a} + \frac{\pi}{2a} = \frac{\pi \ln (a) + \pi}{2a}In summary, we used a
  • #1
alyafey22
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Find the following integral :

\(\displaystyle \int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx\)
 
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  • #2
ZaidAlyafey said:
Find the following integral :

\(\displaystyle \int^{\infty}_{0} \frac{\ln(x) }{x^2+a^2}\,dx\) <----- THERE IT IS!

Hope that helped :)
 
  • #3
ZaidAlyafey said:
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.
 
  • #4
ZaidAlyafey said:
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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  • #5
\(\displaystyle F(b) = \int^{\infty}_0 \frac{x^b}{x^2+a^2}\, \, \) ----(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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  • #6
There is a complex analysis approach , if someone is interested . I will try to post it later .
 
  • #7
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 $0$.

View attachment 923

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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FAQ: Integral of Logarithm: Find & Solve

1. What is the definition of an integral of logarithm?

The integral of logarithm is a mathematical concept that represents the area under the curve of a logarithmic function. It is denoted by ∫log(x)dx and is used to determine the value of a logarithmic function at a given point or to find the area under the curve between two points.

2. How do you solve an integral of logarithm?

To solve an integral of logarithm, you can use integration by parts or substitution. First, you need to rewrite the logarithmic function as a combination of simpler functions. Then, apply the appropriate integration technique to evaluate the integral and find the antiderivative. Finally, use the fundamental theorem of calculus to calculate the definite integral or the value of the function at a specific point.

3. What is the relationship between the derivative and integral of logarithm?

The derivative and integral of logarithm are inverse operations. This means that the derivative of a logarithmic function is equal to the integrand of its integral. In other words, if f(x) = log(x), then f'(x) = 1/x. Similarly, if F(x) = ∫log(x)dx, then F'(x) = log(x).

4. Can you use the properties of logarithms to simplify an integral of logarithm?

Yes, you can use the properties of logarithms, such as the product rule, quotient rule, and power rule, to simplify an integral of logarithm. These properties can help you rewrite the integrand in a more manageable form, making it easier to solve the integral.

5. What are the common applications of integrals of logarithms?

Integrals of logarithms have several applications in mathematics, science, and engineering. They are used to calculate the area under logarithmic curves, which is important in statistics, economics, and physics. They also have applications in differential equations, where they are used to solve problems involving exponential growth and decay. Additionally, integrals of logarithms are used in computer science and data analysis to model and analyze logarithmic data.

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