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?

Originally Posted by MBM

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:

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

Originally Posted by mathmari

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 in to 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)

Originally Posted by ZaidAlyafey

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

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

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

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?