Can you imagine?

Calculus Level 3

What can be stated about the value of the imaginary part of

$\int_0^1 (-1)^x dx?$

Solve using the principal value.

it is

< 0.

it is

> 1.

it is

< 1,

but

> 0.

it is

0.

1 solution

Steven Chase
Jul 14, 2019

$\large{\int_0^1 (-1)^x \, dx = \int_0^1 e^{j \, \pi \, x} \, dx = \frac{1}{j \pi} e^{j \, \pi \, x} \Big|_0^1 = \frac{1}{j \pi} \Big(e^{j \, \pi } - 1 \Big ) = - \frac{2}{j \pi} = \frac{2}{\pi} j }$

Just curious as to why you used “j” and not “i” to represent the imaginary unit?

Δrchish Ray - 1 year, 11 months ago

It's an electrical engineering convention, since "i" is used for current typically

Steven Chase - 1 year, 11 months ago

Understood. Thanks!

Δrchish Ray - 1 year, 11 months ago

How can an imaginary number be compared with real numbers?

A Former Brilliant Member - 1 year, 11 months ago

I interpreted the question as meaning "If the integral is of the form $a + j b$ , what can we say about $b$ ?"

Steven Chase - 1 year, 11 months ago

That's right. But isn't there a mistake in the question itself?

A Former Brilliant Member - 1 year, 11 months ago

@A Former Brilliant Member – The question asks for the imaginary part, which I interpreted as being the scalar multiple in front of the imaginary unit. Perhaps not all readers would interpret it that way though.

Steven Chase - 1 year, 11 months ago

@Steven Chase – Oh, I have overlooked that. Sorry. You are absolutely right.

A Former Brilliant Member - 1 year, 11 months ago

Using "j" to represent an imaginary number is a common practice in Electrical Engineering. It helps us avoid confusion since "i" is taken for current. But then again, "J" also denotes current density!!!

tom engelsman - 1 year, 11 months ago

Can you imagine?

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