If $ I = \int_0^\pi \sin(\sin^{-1}\{x\})dx $ then what is $\big[I\big]$?

Question

I was working on the following question:

If $$ I = \int_0^\pi \sin(\sin^{-1}\{x\})dx $$ then what is $\big[I\big]$? ($ \{x\} $ means the fractional part of $x$)

My solution went like this: $\sin(\arcsin\{x\}) = \{x\}.$ So the integral simply reduces to:

$$ I = \int_0^\pi \{x\}dx $$ $$ I = 3\int_0^1 x dx + \int_0^{ \pi-3 }xdx $$

So, $\big[I\big] = 1$.

But the given solution went like this:

$$ 0 \le \{x\} < 1 $$ $$ 0 \le \arcsin{\{x\}}<\pi/2$$ $$ 0 \le \sin(\arcsin{\{x\})} < 1$$ $$ \int_0^\pi0dx \le \int_0^\pi\sin(\arcsin{\{x\})} < \int_0^\pi1dx$$ $$ \int_0^\pi0dx \le I < \int_0^\pi1$$

So $\big[ I \big] = 3$.

Now, who is wrong? I believe that the given solution is wrong because the inequality used is strict, so even if $3 <\pi$, still $1<\pi$ holds true. But still, $$ 0 \le \sin(\arcsin{\{x\})} < 1$$

which follows from $$ 0 \le \{x\} < 1 $$

is like $\sin(\arcsin{\{x\})}$ takes all values lesser than one, even those like $0.9999...$ So, the solution still makes 'sense', if thought qualitatively.

I like your work better; in fact, I don't understand the given solution. But I could be wrong. — Eric Auld, Jul 30 '13 at 12:49
What is it with the given solution that you don't understand? (The only fault I can find is what I already pointed out) — Parth Thakkar, Jul 30 '13 at 12:51
Have a look at this problem which has some related techniques. — Mhenni Benghorbal, Jul 30 '13 at 12:52
I don't understand why $0 \leq I < \pi$ means that $\lfloor I \rfloor = 3$. — Eric Auld, Jul 30 '13 at 12:53
That's exactly what I was trying to tell! So, maybe the given solution is erroneous? — Parth Thakkar, Jul 30 '13 at 12:54
@EricAuld That's because it doesn't. That only says it could be one of $0,,1,,2,,3$. — Daniel Fischer, Jul 30 '13 at 12:55
The right answer is $$ I = 3\int_0^1 x dx + \int_0^{ \pi-3 }xdx=\frac 3 2 +\frac {(\pi-3)^2} 2 =\frac 1 2 ,{\pi }^{2}-3,\pi +6=1.51002.$$ — user64494, Jul 30 '13 at 13:00
@MhenniBenghorbal, in the problem you told me to see, I don't understand the line $$ = \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{1}{y^3} (1+n)(y-n), dy= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{2} $$. I could've asked there, but asking here would give an instant reply, so. — Parth Thakkar, Jul 30 '13 at 13:02
@user64494 The $[I]$ is the greatest integer part, aka Gauss bracket or floor. — Daniel Fischer, Jul 30 '13 at 13:03
@ParthThakkar: Just evaluate the integral and you will see it. — Mhenni Benghorbal, Jul 30 '13 at 13:09
The inequalities $0 \le I,, I< \pi$ do not imply $[I]=3. $– — user64494, Jul 30 '13 at 13:22

score 2 · Accepted Answer · answered Jul 30 '13 at 14:08

In contrast to the given solution, yours is correct and clear. Maybe a bit brief, adding a "thus $\frac{3}{2} \leqslant I < 2$ between

$$I = 3\int_0^1 x\,dx + \int_0^{\pi-3} x\,dx$$

and "So, $[I] = 1$" would not be detrimental. Depending on what you may consider obvious, adding further intermediate steps could be demanded or superfluous.

In the given solution, from

$$\int_0^\pi 0\,dx \leqslant I < \int_0^\pi 1\, dx$$

one can only deduce $0 \leqslant [I] \leqslant 3$, which of the four possible values is the correct one must be determined by other means. The reported answer $[I] = 3$ is, as shown by your calculation, simply wrong.

If $ I = \int_0^\pi \sin(\sin^{-1}\{x\})dx $ then what is $\big[I\big]$?

1 Answers1