MHB Solving an integral using Differentiation under the integral sign

[MBM1]

I am Trying to solve the difference of the two following integrals:

(1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
(2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
$g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?

 

[mathmari]

I am Trying to solve the difference of the two following integrals:

(1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
(2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
$g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?

You got the difference $\infty - \infty$ and you set it equal to $0$, but this difference is undefined.Using Euler's formula $$e^{ix}=\cos{(x)}+i \sin{(x)}$$
we get $$\cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$$

Therefore we have the following:

$$\int_{0}^{\infty} \frac{cos(kx)}{k}\,dk- \int_{0}^{\infty} \frac{e^{-2k}cos(kx)}{k}\,dk =\int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{-2k}(e^{ikx}+e^{-ikx})}{2k}\,dk= \\ \int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{(ix-2)k}+e^{-(ix+2)k}}{2k}\,dk$$

 

[MBM1]

You got the difference $\infty - \infty$ and you set it equal to $0$, but this difference is undefined.Using Euler's formula $$e^{ix}=\cos{(x)}+i \sin{(x)}$$
we get $$\cos{(x)}=\frac{e^{ix}+e^{-ix}}{2}$$

Therefore we have the following:

$$\int_{0}^{\infty} \frac{cos(kx)}{k}\,dk- \int_{0}^{\infty} \frac{e^{-2k}cos(kx)}{k}\,dk =\int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{-2k}(e^{ikx}+e^{-ikx})}{2k}\,dk= \\ \int_{0}^{\infty} \frac{e^{ikx}+e^{-ikx}}{2k}\,dk -\int_{0}^{\infty} \frac{e^{(ix-2)k}+e^{-(ix+2)k}}{2k}\,dk$$

@ mathmari, thank you for your response. You have decomposed the integral into 4 complex exponential integrals multiplied by a hyperbola however I do not see how it simplifies the integral, i.e. consider the following integral

$\int_{0}^{\infty} \frac{\exp(ikx)}{2k}\,dk$ is still difficult to solve with the knowledge that I have. (or maybe I am not thinking)

 

[chisigma]

I am Trying to solve the difference of the two following integrals:

(1) $g_{1}(x) = \int_{0}^{\infty} \frac{cos(kx)}{k}\,dk$
(2) $g_{2}(x) = \int_{0}^{\infty} \frac{\exp(-2k)cos(kx)}{k}\,dk$

I read the thread on Advanced Integration Techniques and it mentioned the Differentiation under the integral sign technique which I am unfamiliar to. Nonetheless I tried it and found ,
$g_{1}(x)=\infty $ and $g_{2}(x)= \infty + \frac{1}{2}ln(1+\frac{4}{x^2})$

Hence $g_{1}(x) - g_{2}(x) = -\frac{1}{2}ln(1+\frac{4}{x^2})$.

What I would like to know is my final answer true or false and if it is false where did I make an invalid assumption or mistake?

The problem is bad set because both integrals diverge. It is reasonable instead the calculation of the integral...

$\displaystyle g(x) = \int_{0}^{\infty} \frac{1 - e^{-2\ k}}{k}\ \cos k\ x\ dk\ (1)$

... which converges. The calculation of (1) is realatively easy using some advanced technique like the Laplace Transform, much more difficult using an elementary approach...

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

$$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

$$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$

$$F(y)=\frac{\log(1+4y^2)}{2}$$

Hence we have

$$F\left(\frac{1}{y}\right)=\int^\infty_0 \frac{\cos(x)(1-e^{-2\frac{x}{y}})}{x} dx$$

Now let $x/y=t$

$$\int^\infty_0 \frac{\cos(yt)(1-e^{-2t})}{t} dt =\frac{\log\left(1+\frac{4}{y^2} \right)}{2} $$

 

[MBM1]

$$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

$$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$

$$F(y)=\frac{\log(1+4y^2)}{2}$$

Hence we have

$$F\left(\frac{1}{y}\right)=\int^\infty_0 \frac{\cos(x)(1-e^{-2\frac{x}{y}})}{x} dx$$

Now let $x/y=t$

$$\int^\infty_0 \frac{\cos(yt)(1-e^{-2t})}{t} dt =\frac{\log\left(1+\frac{4}{y^2} \right)}{2} $$

thanks a lot. how did you become good in math, do you have any books you could recommend for me to read?

 

[alyafey22]

thanks a lot. how did you become good in math, do you have any books you could recommend for me to read?

You're welcome.

Essentially, my knowledge (of this particular field) is based on reading different articles, papers and books. I would suggest some books but that depends on the area you're interested in.

Since this is not the best place to discuss that , I prefer you present your concerns via visitor messages or private messages.

 

[chisigma]

$$F(y)=\int^\infty_0 \frac{\cos(x)(1-e^{-2xy})}{x} dx$$

$$F'(y)=2\int^\infty_0 e^{-2xy}\cos(x)\,dx = \frac{4y}{1+4y^2}$$

From the formal point of view there is some problem in this step since the rule of derivation under the integral sign is valid only if the range of integration is limited...

Leibniz Integral Rule -- from Wolfram MathWorld

In this particular case, it should be noted that setting y = 0 would lead to the conclusion that ...

$\displaystyle F^{\ '} (0) = \int_{0}^{\infty} \cos x\ dx = 0\ (1)$

which of course is not true...

Kind regards

$\chi$ $\sigma$