Is the sign in this integral correct?

[EngWiPy]

In the table of integral, there is this formula

$$\int_0^{\infty}\frac{sin(ax)}{x}\,dx=\frac{\pi}{2}\text{sign} a$$

is sign a here is literally the sign of a, or it means something else?

 

[Mark44]

In the table of integral, there is this formula

$$\int_0^{\infty}\frac{sin(ax)}{x}\,dx=\frac{\pi}{2}\text{sign} a$$

is sign a here is literally the sign of a, or it means something else?

Yes, it means the sign of a. $\text{sign}(a)$ and equals +1 if a > 0 and -1 if a < 0. So the integral above equals $\pi/2$ if a > 0, and $-\pi/2$ if a < 0.

 

[EngWiPy]

Yes, it means the sign of a. $\text{sign}(a)$ and equals +1 if a > 0 and -1 if a < 0. So the integral above equals $\pi/2$ if a > 0, and $-\pi/2$ if a < 0.

Thanks. I was looking for something similar but for cosine, and I found this

$$\int_0^{\infty}\frac{cos(ax)}{x+\beta}\,dx=-sin(a\beta)si(a\beta)-cos(a\beta)ci(a\beta)$$

where $si(.)$ and $ci(.)$ are the sine and cosine integrals, respectively. However, I want this for $\beta=0$, which results in $-ci(0)$. I am not sure, but it seems the result of this is $-\infty$. Does this mean that $cos(x)/x$ is not integrable over $[0, \infty)$?

 

[Mark44]

Thanks. I was looking for something similar but for cosine, and I found this

$$\int_0^{\infty}\frac{cos(ax)}{x+\beta}\,dx=-sin(a\beta)si(a\beta)-cos(a\beta)ci(a\beta)$$

where $si(.)$ and $ci(.)$ are the sine and cosine integrals, respectively. However, I want this for $\beta=0$, which results in $-ci(0)$. I am not sure, but it seems the result of this is $-\infty$. Does this mean that $cos(x)/x$ is not integrable over $[0, \infty)$?

Right, because the graph of $y = \frac{\cos(x)} x$ has a non-removable discontinuity at 0, unlike the graph of $y = \frac{\sin(x)} x$, which has a removable discontinuity that doesn't affect the integral.

From the graph here, $Ci(0)$ appears to be $-\infty$, so I think you have an incorrect sign.

 

[EngWiPy]

Thanks. Yes, right $ci(0)=-\infty$ not $-ci(0)$. So, I cannot evaluate this integration? The original integration that I want to evaluate is

$$\int_0^{\infty}\frac{e^{-jtx}}{t}\,dt$$

where $j=\sqrt{-1}$. Can I evaluate it in another way?

 

[Svein]

Ahlfors notes that using residues and some limit considerations, the Cauchy principal value of $\int_{-\infty}^{\infty}\frac{e^{jt}}{t}dt$ is $j\pi$. I would suggest substituting $u=x\cdot t$ in your integral and see where it leads.

 

[EngWiPy]

Ahlfors notes that using residues and some limit considerations, the Cauchy principal value of $\int_{-\infty}^{\infty}\frac{e^{jt}}{t}dt$ is $j\pi$. I would suggest substituting $u=x\cdot t$ in your integral and see where it leads.

Thanks, but what about the integration limits? In my case. the lower limit is $0$ and not $-\infty$. Even after the change of variables, the limits remains the same.

 

[Svein]

Thanks, but what about the integration limits? In my case. the lower limit is $0$ and not $-\infty$. Even after the change of variables, the limits remains the same.

I haven't calculated the answer, just presented a tip. Next tip: split the integral in the middle and substitute -u for u in the negative part.

 

[EngWiPy]

I haven't calculated the answer, just presented a tip. Next tip: split the integral in the middle and substitute -u for u in the negative part.

In my case, the exponent is negative, not just the limits. So, if I assume that $u=-t$, I get the following integral:

$$-\int_{-\infty}^{\infty}\frac{e^{-ju}}{u}\,du = -\int_{-\infty}^0\frac{e^{-ju}}{u}\,du -\int_{0}^{\infty}\frac{e^{-ju}}{u}\,du = j\pi$$

I am interested in $\int_{0}^{\infty}\frac{e^{-ju}}{u}\,du = -\int_{-\infty}^0\frac{e^{-ju}}{u}\,du -j\pi$, but then how to evaluate the integral in the RHS. It is the same problem as before.

 

[Svein]

I know that you may still not be satisfied, but $\int_{-\infty}^{0}\frac{e^{-ju}}{u}du=\int_{\infty}^{0}\frac{e^{ju}}{u}du=-\int_{0}^{\infty}\frac{e^{ju}}{u}du$. Add the two integrals: $\int_{0}^{\infty}\frac{e^{-ju}}{u}du-\int_{0}^{\infty}\frac{e^{ju}}{u}du=2\cdot j \int_{0}^{\infty}\frac{\sin(u)}{u}du$.