# Improper integral

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• Dec 16th 2011, 05:52 AM
fareastmovement
Improper integral
Derive the integration formula
$\displaystyle \int_0^{\infty}\frac{cos(ax)-cos(bx)}{x^2}dx=\frac{\pi}{2}(b-a)$

By using the residue theorem, I integrated the function $\displaystyle \frac{e^iaz-e^ibz}{z^2}$ from $\displaystyle -\infty$to $\displaystyle \infty$, I can only obtain the result $\displaystyle 2\pi i(b-a)$, which means my answer is $\displaystyle \pi(b-a)$.
I couldnt find my mistake.Thank you.
• Dec 16th 2011, 06:44 AM
girdav
Re: Improper integral
The function you have to integrate is even, so the integral on $\displaystyle \mathbb R$ is two times the integral on $\displaystyle \mathbb R_+$.
• Dec 16th 2011, 07:10 AM
fareastmovement
Re: Improper integral
Yes i did it (divided by 2)and still get $\displaystyle \pi(b-a)$ as answer, but not$\displaystyle \frac{\pi}{2}(b-a)$
• Dec 16th 2011, 07:23 AM
girdav
Re: Improper integral
Did you wrote the exponentials in terms of $\displaystyle \cos$?
• Dec 16th 2011, 07:50 AM
fareastmovement
Re: Improper integral
Yes, by integrating $\displaystyle \frac{e^{iaz}-e^{ibz}}{z^2}$ and compare the real part of both sides
• Dec 17th 2011, 08:41 AM
FernandoRevilla
Re: Improper integral
Another way: denote $\displaystyle I(\lambda)=\int_0^{+\infty}\frac{\cos \lambda x}{x^2}dx\;(\lambda >0)$ , then $\displaystyle I'(a)=-\int_0^{+\infty}\frac{\sin \lambda x}{x}dx$ . Using the substitution $\displaystyle t=\lambda x$ we get $\displaystyle I'(\lambda)=-\int_0^{+\infty}\frac{\sin t}{t}dt=-\frac{\pi}{2}$ (Dirichlet's integral) . So, $\displaystyle I(\lambda)=-\frac{\pi}{2}\lambda+C$ . For $\displaystyle \lambda =0$ we get $\displaystyle C=\int_0^{+\infty}\frac{dx}{x^2}$ .

Then,

$\displaystyle \int_0^{+\infty}\frac{\cos ax-\cos bx}{x^2}dx=-\frac{\pi}{2}a+C-\left(-\frac{\pi}{2}b+C\right)=\boxed{\frac{\pi (b-a)}{2}}$

@fareastmovement: Show your complete work by the method of residues (of course if you want) and we can check it.

Edited: Method not valid, see the posts below.
• Dec 17th 2011, 09:30 AM
fareastmovement
Re: Improper integral
Thank you, I got it now. You are really helpful :)
• Dec 18th 2011, 10:25 PM
chisigma
Re: Improper integral
Quote:

Originally Posted by FernandoRevilla
Another way: denote $\displaystyle I(\lambda)=\int_0^{+\infty}\frac{\cos \lambda x}{x^2}dx\;(\lambda >0)$ , then $\displaystyle I'(a)=-\int_0^{+\infty}\frac{\sin \lambda x}{x}dx$ . Using the substitution $\displaystyle t=\lambda x$ we get $\displaystyle I'(\lambda)=-\int_0^{+\infty}\frac{\sin t}{t}dt=-\frac{\pi}{2}$ (Dirichlet's integral) . So, $\displaystyle I(\lambda)=-\frac{\pi}{2}\lambda+C$ . For $\displaystyle \lambda =0$ we get $\displaystyle C=\int_0^{+\infty}\frac{dx}{x^2}$ .

Then,

$\displaystyle \int_0^{+\infty}\frac{\cos ax-\cos bx}{x^2}dx=-\frac{\pi}{2}a+C-\left(-\frac{\pi}{2}b+C\right)=\boxed{\frac{\pi (b-a)}{2}}$

@fareastmovement: Show your complete work by the method of residues (of course if you want) and we can check it.

There is only a minor detail: writing $\displaystyle C=\int_{0}^{+ \infty} \frac{dx}{x^{2}}$ is a nonsense because the integral diverges...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 18th 2011, 11:00 PM
chisigma
Re: Improper integral
In order to avoid criticity in the solution of the integral I would suggest to use the trigonometric identity...

$\displaystyle \cos \alpha - \cos \beta = 2\ \sin \frac{\beta + \alpha}{2}\ \sin \frac{\beta-\alpha}{2}$ (1)

... so that the integral becomes...

$\displaystyle I(a,b)=2\ \int_{0}^{\infty} \frac{\sin (\frac{b+a}{2}\ x)\ \sin (\frac{b-a}{2}\ x)}{x^{2}}\ dx$

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 18th 2011, 11:01 PM
FernandoRevilla
Re: Improper integral
Quote:

Originally Posted by chisigma
There is only a minor detail: writing $\displaystyle C=\int_{0}^{+ \infty} \frac{dx}{x^{2}}$ is a nonsense because the integral diverges...

Right, just a lapse of concentration. Easily avoided: $\displaystyle I(1/2)$ is finite, so $\displaystyle I(1/2)=-\pi/4+C$ and $\displaystyle C=I(1/2)+\pi/4$ is finite.
• Dec 18th 2011, 11:18 PM
chisigma
Re: Improper integral
Quote:

Originally Posted by FernandoRevilla
Right, just a lapse of concentration. Easily avoided: $\displaystyle I(1/2)$ is finite, so $\displaystyle I(1/2)=-\pi/4+C$ and $\displaystyle C=I(1/2)+\pi/4$ is finite.

The real problem is in defining the function ...

$\displaystyle I(\lambda)= \int_{0}^{\infty} \frac{\cos (\lambda x)}{x^{2}}\ dx$ (2)

... because the integral diverges for any value of $\displaystyle \lambda$. In my previous post a way to overcame such type of criticity is indicated...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 18th 2011, 11:43 PM
FernandoRevilla
Re: Improper integral
Quote:

Originally Posted by chisigma
... because the integral diverges for any value of $\displaystyle \lambda$. In my previous post a way to overcame such type of criticity is indicated...

Not for all values of $\displaystyle \lambda$ but of course the method I used is not valid. Reviewing the situation I find the explanation: in a foolish way my mind thought about $\displaystyle 1$ to $\displaystyle \infty$ instead of $\displaystyle 0$ to $\displaystyle \infty$ and curiously at the end I made the substitution $\displaystyle \lambda=0$ . A complete disaster. :)

Method not valid.
• Dec 19th 2011, 01:39 AM
Also sprach Zarathustra
Re: Improper integral
Quote:

Originally Posted by fareastmovement
Derive the integration formula
$\displaystyle \int_0^{\infty}\frac{cos(ax)-cos(bx)}{x^2}dx=\frac{\pi}{2}(b-a)$

By using the residue theorem, I integrated the function $\displaystyle \frac{e^iaz-e^ibz}{z^2}$ from $\displaystyle -\infty$to $\displaystyle \infty$, I can only obtain the result $\displaystyle 2\pi i(b-a)$, which means my answer is $\displaystyle \pi(b-a)$.
I couldnt find my mistake.Thank you.

Perhaps, using integration by parts and aplaying Furlani's integrals we can conclude something.

$\displaystyle \int_0^{\infty} \frac{\cos{ax}-\cos{bx}}{x^2}dx=\frac{\cos{bx}-\cos{ax}}{x}\big{|}_{\mathbb{R^+}$$\displaystyle + \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx$

Now, using the theorem:

$\displaystyle \int_0^{\infty} \frac{f(ax)-f(bx)}{x} dx=\(f(0)-f(\infty))\ln\frac{b}{a}$

I stuck on evaluating:

$\displaystyle \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx$

Edit:

Ok... No Furalani needed.

Using only the well known Dirichlet's integral we get:

$\displaystyle \int_0^{\infty} \frac{b\sin{bx}-a\sin{ax}}{x} dx=\frac{\pi}{2}(b-a)$

(While $\displaystyle \frac{\cos{bx}-\cos{ax}}{x}\big{|}_{\mathbb{R^+}$ $\displaystyle {=0}$)
• Dec 19th 2011, 03:34 AM
FernandoRevilla
Re: Improper integral
(a) $\displaystyle I(a,b)=\int_0^{\infty}\frac{\cos ax-\cos bx}{x^2}dx$ is convergent for $\displaystyle a>0,b>0$ (easily proved using chisigma's answer #9 )

(b) $\displaystyle \frac{\partial I}{\partial a}=\ldots=-\frac{\pi}{2},\;\frac{\partial I}{\partial b}=\ldots=\frac{\pi}{2}$ (differentiation under the integral sign and Dirichlet's Integral).

(c) $\displaystyle I(a,b)=-\frac{\pi}{2}a+\varphi(b)$ , so $\displaystyle \frac{\partial I}{\partial b}=\varphi'(b)=\frac{\pi}{2}$ hence $\displaystyle \varphi(b)=\frac{\pi}{2}b+C$ . That is, $\displaystyle I(a,b)=\frac{\pi (b-a)}{2}+C\quad (*)$ .

(d) Taking limits in $\displaystyle (*)$ as $\displaystyle (a,b)\to (0,0)$ we get $\displaystyle C=0$ . So, $\displaystyle I(a,b)=\frac{\pi (b-a)}{2}$
• Dec 19th 2011, 11:19 AM
Random Variable
Re: Improper integral
$\displaystyle \int^{\infty}_{0} \frac{\cos ax - \cos bx}{x^{2}} \ dx$

$\displaystyle = \int_{0}^{\infty} \int_{a}^{b} \frac{\sin tx}{x} \ dt \ dx$

Fubini's theorem is not satsified. So to justify changing the order of integration, let's write the double integral more formally as a limit.

$\displaystyle = \lim_{n \to \infty} \int_{0}^{n} \int_{a}^{b} \frac{\sin tx}{x} \ dt \ dx = \lim_{n \to \infty} \int_{a}^{b} \int_{0}^{n} \frac{\sin tx}{x} \ dx \ dt$

$\displaystyle = \lim_{n \to \infty} \int_{a}^{b} \int_{0}^{nt} \frac{\sin u}{u} \ du \ dt = \lim_{n \to \infty} \int_{a}^{b} \text{Si} (nt) \ dt$

Since the Sine Integral is bounded above by 2, we can use the Dominated Convergence Theorem to justify bringing the limit inside of the integral.

$\displaystyle = \frac{\pi}{2} \int_{a}^{b} \ dt = \frac{\pi}{2} (b-a)$
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