Need help with (simple) math problem (convergence of infinit integra).

[Ratamahatta]

Hello, I need help with this problem. I don't want solution (not yet anyway [img]/images/graemlins/smile.gif[/img] ), I just want some one to give me a hint.

I need to show that this integral:

int_(from 0 to a) 1/[sin(sqrt(x))] dx 0 < a < pi.

is convergent.

I know that I need to put some constant (h) in stead of 0 and then find lim_h \to 0, but I can't solve this integral. I tried substitution, but it gets very messy. I even tried writing Riemann-sum for that function and rewriting 1/sin(sqrt(x)) as a series but I had to stop because it got to complicated.

So, does anyone have any hints? Thanks!

[ADDboy]

I would do this by comparison. Find a function that is larger than 1[sin(sqrt(x))] for x close to 0 yet whose integral still converges. As a hint for how to find such a function, think about the Taylor series expansion for sin(x).

[ncray]

There are many ways to do this problem. I think the easiest might be to let g(x) = 1/sin(sqrt(pi)), which will be a constant function, and evaluate the integral of g from 0 to pi (just pi/sin(sqrt(pi)). You need to show that g(x) >= your original function at every point on the interval.

[Siegmund]

ncray: only problem there is that g(x) < original function for small x. The original function shoots up to infinitity near x=0...

But all is not lost, OP. Remember that lim (sin x)/x is 1 as x approaches 0? That means lim (sin sqrt(x))/sqrt(x) is too.

[ncray]

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ncray: only problem there is that g(x) < original function for small x. The original function shoots up to infinitity near x=0...

But all is not lost, OP. Remember that lim (sin x)/x is 1 as x approaches 0? That means lim (sin sqrt(x))/sqrt(x) is too.

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Right, my bad. I was thinking of cosine.

[Ratamahatta]

[ QUOTE ]
I would do this by comparison. Find a function that is larger than 1[sin(sqrt(x))] for x close to 0 yet whose integral still converges. As a hint for how to find such a function, think about the Taylor series expansion for sin(x).

[/ QUOTE ]

I thought about doing it by comparison but couldn't find such a funktion.
Buy Taylor series expansion for sin(x) you mean: x - (x^3)/(3!) + (x^5)/(5!) (x^7)/(7!)...? I can't really see how I can use it though especialy because it's an alternating series.

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But all is not lost, OP. Remember that lim (sin x)/x is 1 as x approaches 0? That means lim (sin sqrt(x))/sqrt(x) is too.


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Same here... can't really see how it can be used. [img]/images/graemlins/confused.gif[/img]

[Ratamahatta]

[ QUOTE ]


[ QUOTE ]
But all is not lost, OP. Remember that lim (sin x)/x is 1 as x approaches 0? That means lim (sin sqrt(x))/sqrt(x) is too.


[/ QUOTE ]

Same here... can't really see how it can be used. [img]/images/graemlins/confused.gif[/img]

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Ah wait a minute! Do you mean that 1/sin(sqrt(x)) can be compared to 1/sqrt(x)? sin(sqrt(x))/sqrt(x) < 1, therefor sin(sqrt(x)) < sqrt(x) which makes 1/sin(sqrt(x)) > 1/sqrt(x) and in order to compare them we need another function to be higher than 1/sin(sqrt(x)).