# Proof: Compare two integral(Please look at my surgested proof)

• Hummingbird25
In summary, the suggested proof compares two integrals and provides a concise explanation of their similarities and differences. It explains the concept of integration and how it is used to find the area under a curve. The proof also highlights the importance of using appropriate techniques and notation in integration. Additionally, it discusses the concept of indefinite and definite integrals and how they differ in terms of their limits of integration. Overall, the proof offers a clear and concise comparison of two integral methods and provides a better understanding of integration in mathematics.
Hummingbird25
This is a repost, the reason I feared that people who miss the the original.

## Homework Statement

Looking at the Integral

$$a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}$$

prove that $$a_n \geq a_{n+1}$$

## The Attempt at a Solution

Here is my now proof:

the difference between the two integrals, we seek to show:

$$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0$$

Common denominator:

$$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt$$

$$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt$$

From here we can use the fact that the denominator on the half of the interval where sinus is negative is larger than the denominator of each and every other corresponding point on the other half of the interval, so the whole integral must be positive.

q.e.d.

How does it look now?

Sincerely Maria.

But sin(t) isn't negative anywhere on [0,pi]. That makes life a lot easier.

Looks good.

Hi Dick,

So I change my conclusion.

Since sin(t) is positive on the interval [0,pi] then the who integral most be positive and thusly convergent.

Does this sound better?

Sincerely

Maria

If everything is nonnegative all you need is that t+n*pi<t+(n+1)*pi. So 1/(t+n*pi)>1/(t+(n+1)*pi). It becomes pretty obvious that your integral is positive.

Dick said:
If everything is nonnegative all you need is that t+n*pi<t+(n+1)*pi. So 1/(t+n*pi)>1/(t+(n+1)*pi). It becomes pretty obvious that your integral is positive.

And thusly it converges?

Sincerely
Maria.

Hummingbird25 said:
And thusly it converges?

Sincerely
Maria.

It converges because the integrand and domain of integration are bounded, right?

Dick said:
It converges because the integrand and domain of integration are bounded, right?

Here is my original proof:

Looking at the Integral

$$a_n = \int_{0}^{\pi} \frac{sin(x)}{x+n\pi}$$

prove that $$a_n \geq a_{n+1}$$

## The Attempt at a Solution

Here is my now proof:

the difference between the two integrals, we seek to show:

$$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0$$

Common denominator:

$$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt$$

$$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt$$

The integral is therefore non-negativ and therefore

$$t+n*pi<t+(n+1)*pi$$ and $$/(t+n*pi)>1/(t+(n+1)*pi).$$ thus the integral is positive.

The integral is therefore bounded since the integrand is continuous which makes the original in-equality true.

q.e.d.

Is it on the mark now?

Sincerely Yours
Maria.

i) Your 'therefores' are somewhat backwards. sin(t)/(t+n*pi)>sin(t)/(t+(n+1)*pi) implies the integrand is positive. The integrand being positive DOESN'T imply the inequality. So, ii) you don't NEED a common denominator anymore. All the parts are there, you could make the phrasing clearer.

Dick said:
i) Your 'therefores' are somewhat backwards. sin(t)/(t+n*pi)>sin(t)/(t+(n+1)*pi) implies the integrand is positive. The integrand being positive DOESN'T imply the inequality. So, ii) you don't NEED a common denominator anymore. All the parts are there, you could make the phrasing clearer.

Here we go again:

proof:

By applying the limit

$$sin(t)/(t+n*pi)>sin(t)/(t+(n+1)*pi)$$ the integrand becomes positive for all n. The integral is positive and inequality is thusly true.

Best Regards.
Maria.

It's not a 'limit', it's an inequality. But good enough.

Last edited:

## 1. What is the purpose of comparing two integrals?

The purpose of comparing two integrals is to determine if they have the same value or if one is larger or smaller than the other. This can help in evaluating the convergence or divergence of a series or in finding the area under a curve.

## 2. How do you compare two integrals?

To compare two integrals, you can use various methods such as the comparison test, limit comparison test, or direct comparison test. These tests involve comparing the given integral to a known integral that is either larger or smaller, and then using their properties to determine the convergence or divergence of the given integral.

## 3. Can you explain the process of using the comparison test to compare two integrals?

To use the comparison test, you first need to find a known integral that is either larger or smaller than the given integral. Then, you compare the two integrals and use their properties to determine if the given integral is convergent or divergent. If the known integral is convergent, then the given integral must also be convergent. If the known integral is divergent, then the given integral must also be divergent.

## 4. Are there any other methods for comparing two integrals?

Yes, there are other methods for comparing two integrals such as the ratio test and the root test. These tests involve taking the limit of the ratio or the root of the given integral and comparing it to a known value to determine the convergence or divergence of the given integral.

## 5. Can you provide an example of using the comparison test to compare two integrals?

For example, if we want to compare the integral ∫1 1/x dx to the integral ∫1 1/x2 dx, we can use the comparison test by first noting that 1/x < 1/x2 for all x ≥ 1. Then, we can integrate both integrals and see that ∫1 1/x dx = ln(x)|1 = ∞ and ∫1 1/x2 dx = -1/x|1 = 1. Since the known integral ∫1 1/x2 dx is convergent, the given integral ∫1 1/x dx must also be convergent.

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