# Thread: help with other integrals??

1. ## help with other integrals??

I want to check if my answers is correct??

if somebody can help??

2. I shall do,
$\int_0^{\infty} \frac{\sin x}{x} dx$

This is both a type I and II improper integral. It is a type II because it is not-defined at a point.

We can subdivide the region into as follows,
$\int_0^1 \frac{\sin x}{x}dx+\int_1^{\infty} \frac{\sin x}{x} dx$

If both integral exists, then the integral above exists as well.
Let us do the second summand,
$\int_1^{\infty} \frac{\sin x}{x}dx$.

We shall find the anti-derivative (or at least try) for,
$\int \frac{\sin x}{x}dx=\int \frac{1}{x}\cdot \sin xdx$
Let,
$u=\frac{1}{x}$ and $v'=\sin x$.
Thus,
$u'=-\frac{1}{x^2}$ and $v=-\cos x$.
Thus,
$-\frac{\cos x}{x}-\int \frac{\cos x}{x^2} dx$
When we take the limit we have,
$-\frac{\cos x}{x}\big|_1^{\infty}-\int_1^{\infty} \frac{\cos x}{x^2}dx$
The first summand certainly exists, (I hope you see why).
The second summand can be "squeezed",
$-\frac{1}{x^2}\leq \frac{\cos x}{x^2} \leq \frac{1}{x^2}$
Thus, it is suffienct (for convergence) for,
$-\int_1^{\infty} \frac{1}{x^2}dx\leq \int_1^{\infty} \frac{\cos x}{x^2} dx\leq \int_1^{\infty} \frac{1}{x^2}dx$
Those, two integral converge (p-series for p=2>1).
Thus,
$\int_1^{\infty} \frac{\cos x}{x^2}$ exists.

Now the question remains to show,
$\lim_{n\to 0^+}\int_n^1 \frac{\sin x}{x}dx$
Expand this in the Taylor series,
$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-...$
Integrate, term by term,
$x-\frac{x^3}{3\cdot 3!}+...$
The substituting $n$ and then taking $n\to 0^+$ will result in 0. And substituting 1 will result in an infinite sum,
$1-\frac{1}{3\cdot 3!}+...$
Which will certainly converge, because the terms are alternating, decreasing, and their limit is zero. Leibnize alternating test gaurenttes convergence.

3. By similar reasoning we can subdivide.
Thus, we need to show (it is sufficient) for the integral,
$\int_1^{\infty}\left| \frac{\sin x}{x} \right| dx$
To diverge, by the Cauchy-Swartzh Inequality,
$\int_1^{\infty} \left| \frac{\sin x}{x} \right|dx \geq \sqrt{ \int_1^{\infty} |\sin x |dx \cdot \int_1^{\infty} \frac{1}{x} dx}$
Now, the absolute value is important.
It tells us that,
$\int_1^{\infty}|\sin x|dx\to \infty$
(And not by oscillation divergence).
And also we know that,
$\int_1^{\infty}\frac{1}{x}dx$
Diverges.
Thus,
We have a function that is larger than divergent function to infinity. Thus it must diverge.

(I am not sure abour the inequality for non-closed intervals).

4. Does thefollowing integral converge:

$\int_1^5\frac{dx}{\sqrt{x^4-1}}$

Informal approach:

Factorise the integrand:

$\int_1^5\frac{dx}{\sqrt{x-1} \sqrt{x+1}\sqrt{x^2+1}}$

The only term that behaves badly in the integration range is the first of the
square roots. So we only need to worry about the behavior of the integrand
near $x=1$. Near $x=1$ the integrand behaves like:

$\frac{1}{2\sqrt{x-1}}$

Which is integrable on $[1,a]$, so the integral does converge.

RonL

5. CaptainBlack is right and I believe the second one converges too (only possible problem at x = pi but the limit is 0).

Nice: the sin(x)/x, of which ThePerfectHacker showed it converges, converges to pi/2

The last problem is a nice one. First we show that:

$
\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\cos x} \right)} dx
$

Indeed, letting t = pi/2-x (limits change) thus dt = -dx or dt = dx and the limits don't change.

Now we start with twice the integral and write it as the sum of the integrals above:

$
\begin{array}{*{20}l}
{2\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx} & { = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx + \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\cos x} \right)} dx} \\
{} & { = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x\cos x} \right)} dx} \\
{} & { = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\frac{{\sin 2x}}{2}} \right)} dx} \\
\end{array}
$

We split this last integral by using a property of the logarithm:

$
\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin 2x} \right)} dx - \int\limits_0^{\frac{\pi }{2}} {\ln 2} dx = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin 2x} \right)} dx - \frac{\pi }{2}\ln 2
$

Notice that the last term is exactly the answer we want! But since we're computing twice the initial integral, we'll want that the remaining integral is equal to this answer as well. Let's go over to sin(t) by letting t = 2x:

$
\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin 2x} \right)} dx = \frac{1}{2}\int\limits_0^\pi {\ln \left( {\sin t} \right)} dt = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin t} \right)} dt
$

The first equality is the substitution and I claim the second is correct. Indeed, sin(x) is symmetrical arround x = pi/2 so that integral from 0 to pi/2 is the same as the one from pi/2 to pi. Then obviously, the one going from 0 to pi is twice that value, together with the factor 1/2 we have the initial integral. I'll put it together (letting t = x again):

$
\begin{array}{*{20}c}
{2\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx} & { = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin 2x} \right)} dx - \frac{\pi }{2}\ln 2} \\
{} & { = \int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx - \frac{\pi }{2}\ln 2} \\
\end{array}
$

Conclusion:

$
{\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\sin x} \right)} dx = - \frac{\pi }{2}\ln 2}
$

6. Good job, TD!

There is just one error, you made the assumption that the integral exists on this interval! If that is true then what you said is true. If that is false then what you said is false.

7. I think the existence is rarely verified when the problem states "compute the integral" with a given answer.
It depends how 'deep' you want to go of course, I wouldn't call it an error but an implicit assumption

$\int_1^{\infty} \left| \frac{\sin x}{x} \right| dx$
$\sum_{n=1}^{\infty} \left| \frac{\sin n}{n} \right|$.