Finding the Limit of a Definite Integral in an Integral Problem

[Saitama]

Too complicated!

There's no need for the change of variables here.

All that is needed is a relationship between $f(r) \equiv \int_0^{\pi/2} x^r \sin x\,dx$ and $g(r) \equiv \int_0^{\pi/2} x^r \cos x\,dx$. Integration by parts will give that relationship. It's best to choose u and v such that $uv\bigl|_0^{\pi/2} = 0$.

Hi D H! :)

I used integration by parts and got the following relations:

$$f(r)=rg(r-1)$$
$$g(r)=\frac{f(r+1)}{r+1}$$

How should I use the above?

 

[D H]

Hi D H! :)

I used integration by parts and got the following relations:

$$f(r)=rg(r-1)$$
$$g(r)=\frac{f(r+1)}{r+1}$$

Very good.

How should I use the above?

For one thing, you can use it to show that Q and R are the same question. If you can solve one you can solve the other.

 

[Saitama]

For one thing, you can use it to show that Q and R are the same question. If you can solve one you can solve the other.

D H, I already solved Q and R, I still haven't been able to find the correct approach for P. Can you please look at my post #34? I used integration by parts as haruspex suggested and now I think the limit goes to zero for the new integral (as $r \rightarrow \infty$ and $0 < 1-x < 1$). Is that correct?

 

[D H]

Suppose you ignore the sin(x) term in the integral: $\int_0^{\pi/2} x^r dr$ . Since 0 < sin(x) < 1 for all x in (0,pi/2), this gives an upper bound on f(r). Substituting that sin(x) term with sin(1) leads to another simple integral. Show that this is a lower bound on f(r).

Now use these bounds to show that $f(r)=c_r \int_0^{\pi/2} x^r dr$ where c_r is some value between sin(1) and 1. What is the behavior of c_r as r→∞?