- #1

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

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

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- Thread starter EngWiPy
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- #1

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

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

- #2

Mark44

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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.

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

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

- #3

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- 61

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

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

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)##?

- #4

Mark44

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

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

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)##?

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

- #5

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[tex]\int_0^{\infty}\frac{e^{-jtx}}{t}\,dt[/tex]

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

- #6

Svein

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- #7

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Cauchy principal valueof [itex]\int_{-\infty}^{\infty}\frac{e^{jt}}{t}dt [/itex] is [itex]j\pi [/itex]. I would suggest substituting [itex]u=x\cdot t [/itex] 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.

- #8

Svein

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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.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.

- #9

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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:

[tex]-\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[/tex]

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.

- #10

Svein

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