# Homework Help: Confusing Integration Question?

1. Aug 5, 2012

### unscientific

Confusing Integration Question??

1. The problem statement, all variables and given/known data

The question is as attached.

3. 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...

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2. Aug 5, 2012

### Ray Vickson

Re: Confusing Integration Question??

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

RGV

3. Aug 5, 2012

### unscientific

Re: Confusing Integration Question??

But when they take the limit of y->0, all you get is I(0) = ∫ sinx /x dx = tan-10 = 0

When you take the limit of y-> ∞, all you get is I(∞) = 0 (If you substitute ∞ into the question)

Last edited: Aug 5, 2012
4. Aug 5, 2012

### voko

Re: Confusing Integration Question??

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

5. Aug 5, 2012

### unscientific

Re: Confusing Integration Question??

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!!