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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
Mentor
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
Mentor
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.

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.

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.

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: $\int_{0}^{\infty}\frac{e^{-ju}}{u}du-\int_{0}^{\infty}\frac{e^{ju}}{u}du=2\cdot j \int_{0}^{\infty}\frac{\sin(u)}{u}du$.