Confusing Integration Question?

[unscientific]

Confusing Integration Question??

Homework Statement

The question is as attached.

The Attempt at a Solution

I am very confused by the question. Firstly is I_(y) a function like (tan^-1y) or an operator like ∫^y?

Then they say I_(y) - I_(∞) = (∏/2) - tan^-1y so I suppose that I_(y) is basically just -tan^-1y.I_(y) - I_(∞) = (∏/2) - tan^-1y

lim_(y→0) [ I_(y) - I_(∞) ] = (∏/2) - 0 = (∏/2)

But returning to the question, as you let y→0, all you get is

y₍₀₎ = ∫ sinx /x dx = tan^-10 = 0

But that doesn't look right at all.I hope someone can clear up all the terminology they are using...

 

[Ray Vickson]

Homework Statement

The question is as attached.

The Attempt at a Solution

I am very confused by the question. Firstly is I_(y) a function like (tan^-1y) or an operator like ∫^y?

Then they say I_(y) - I_(∞) = (∏/2) - tan^-1y so I suppose that I_(y) is basically just -tan^-1y.


I_(y) - I_(∞) = (∏/2) - tan^-1y

lim_(y→0) [ I_(y) - I_(∞) ] = (∏/2) - 0 = (∏/2)

But returning to the question, as you let y→0, all you get is

y₍₀₎ = ∫ sinx /x dx = tan^-10 = 0

But that doesn't look right at all.


I hope someone can clear up all the terminology they are using...

The terminology is very clear: I(y) is a function of y, given by a formula in the picture.

RGV

 

[unscientific]

But when they take the limit of y->0, all you get is I(0) = ∫ sinx /x dx = tan^-10 = 0When you take the limit of y-> ∞, all you get is I(∞) = 0 (If you substitute ∞ into the question)

 

[voko]

Then they say I_(y) - I_(∞) = (∏/2) - tan^-1y so I suppose that I_(y) is basically just -tan^-1y.

They say very plainly that I(\infty) = 0, so I(y) = \frac {\pi} {2} - tan^{-1}(y), including y = 0.

 

[unscientific]

They say very plainly that I(\infty) = 0, so I(y) = \frac {\pi} {2} - tan^{-1}(y), including y = 0.

Oh I think I got it now! Cause when you integrate ∂/∂y = - 1/(1+y²) you get

I_(y) = -tan^-1y + C

Based on the first equation, we know y(∞) but not y(0).

I_(∞) = 0, so you solve C = ∏/2...


Everything makes sense now, thank you!