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Which of the following improper integrals is (are) convergent?

  1. $\int ^{1} _{0} \frac{\sin x}{1-\cos x}dx$
  2. $\int ^{\infty} _{0} \frac{\cos x}{1+x} dx$
  3. $\int ^{\infty} _{0} \frac{x}{1+x^2} dx$
  4. $\int ^{1} _{0} \frac{1-\cos x}{\frac{x^5}{2}} dx$
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4 Comments

Ma'am convergence and divergence is totally different topic in integration. Can we expect questions on them?
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May come for 1 marks.

First we need to do integration , and next check if it converges or diverges , right??
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So @srestha

if the answer is definite then it converges...

and for divergence ?

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1 Answer

3 votes
3 votes

Some theory :-

$p-$test for discontinuity :-. For $a>0$ ,

  1. If $p<1$, then $\large \int_{0}^{a}\frac{1}{x^{p}}dx$ converges .
  2. If $p \geq 1$, then $\large \int_{0}^{a}\frac{1}{x^{p}}dx$ diverges .

$p-$test for Infinite limits :-.  For $a>0$ ;

  1. If $p>1$, then $\large \int_{a}^{\infty}\frac{1}{x^{p}}dx$ converges .
  2. If $p \leq 1$, then $\large \int_{a}^{\infty}\frac{1}{x^{p}}dx$ diverges .

Option 1:-

$\large \int_{0}^{1}\frac{\sin x }{1-\cos x} dx$

let , $\large z=1-\cos x$

    $\large dz=\sin x dx$

Now our integral looks like ,

$\large \int_{0}^{0.459}\frac{1}{z}dz$ 

Now if we compare it with $p-$test for discontinuity 2nd part , here p=1 a>0 which match the condition .

So this improper integral diverges .


Option 3:-

$\large \int_{0}^{\infty}\frac{x}{1+x^{2}}dx$

Let ,$z=1+x^2$

       $dz=2x dx$

Now our integral looks like ,

$\large \frac{1}{2}\int_{1}^{\infty}\frac{1}{z}dz$

If we compare it with $p-$test for Infinite limits , here $p=1$ ,$a>0$ , It matches with the 2nd one .

So this improper integral also diverges.


Option 4:-

$\large 2*\int_{0}^{1}\frac{1-\cos x}{x^{5}}dx$

$I$=$\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx-2\int_{0}^{1}\frac{\cos x}{x^{5}}dx$

$I$=$I_{1}-I_{2}$ 

where $I_{1}=$$\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx$ ,$I_{2}=$ $\large 2\int_{0}^{1}\frac{\cos x}{x^{5}}dx$

Now , $\large 2*\int_{0}^{1}\frac{1}{x^{5}}dx$ this integral diverges as it is matches $p-$test for discontinuity case 2 as here p=5 >1 and a=1>0 .

So as $I_{1}$ diverges the whole improper integral $I$ also diverges . So 

This is also diverges .


Only correct option is (B) It converges .

 Reference

4 Comments

@ankitgupta.1729 sir can you give a efficient explanation for option B i am not getting any easy way for that.

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edited by

If the improper integral $\int_{0}^{\infty} \frac{\cos x}{x} \ dx$ has a finite value then we can say it is a convergent integral.

Since, $1+x \geq \sqrt{x}$ for all $x \geq 0$ which implies $\frac{1}{\sqrt{x}} \geq \frac{1}{1+x}$ for all $x > 0$ $\ (1)$

And $-1 \ \leq \cos x \leq  +1$ $\ \ (2)$

From (1) and (2), we conclude that

$\left | \frac{\cos x}{1+x} \right | \leq \ \left | \frac{\cos x}{\sqrt{x}} \right |$ And so,

 $$\int_{0}^{\infty}\left | \frac{\cos x}{1+x} \right | dx \leq \ \int_{0}^{\infty} \left | \frac{\cos x}{\sqrt{x}} \right | dx$$

You can visualize it here: https://www.desmos.com/calculator/m30qzeiuam

You can see that graph of $\frac{\cos x}{\sqrt{x}}$ covers the graph of $\frac{\cos x}{1+x}$ from both sides i.e. above the axis and below the axis for the interval $(0,\infty)$ So, total area for $\frac{\cos x}{1+x}$ will always be less than the total area for $\frac{\cos x}{\sqrt{x}}$ for the given interval.

Hence, if the value of  $\ \int_{0}^{\infty}  \frac{\cos x}{\sqrt{x}}  dx$ is finite then the integral $\int_{0}^{\infty} \frac{\cos x}{1+x} \ dx$ will be less than a finite value and hence, it will be a convergent integral.

Claim : $\ \int_{0}^{\infty}  \frac{\cos x}{\sqrt{x}}  dx = \sqrt{\frac{\pi}{2}}$

Can you prove it ?

Edit 1: Missed '1' in $\int_{0}^{\infty} \frac{\cos x}{1+x} \ dx$ at the end, so corrected it. 

Edit 2: Both $\frac{\cos x}{1+x}$ and $\frac{\cos x}{\sqrt{x}}$ are continuous in the given interval and $\frac{\cos x}{1+x}$ is defined at zero.

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$\large \int_{0}^{\infty}\frac{\cos x}{\sqrt{x}} dx$

Let , $\large z=\sqrt{x}$

     $\large dz=\frac{1}{2\sqrt{x}}dx$

     $\large 2dz=\frac{1}{\sqrt{x}}dx$

so .$\large 2\int_{0}^{\infty}\cos z^{2}dz$

Now ,

.$\large \int_{0}^{\infty}\cos z^{2}dz$=$\large \frac{\pi}{\sqrt{8}}$  [ from fresnel integral]

So, our desired result ,

$\large \int_{0}^{\infty}\frac{\cos x}{\sqrt{x}} dx$ =$\large 2*\frac{\pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}$

Ok so from this we conclude option (B) converges.

@ankitgupta.1729 thanks sir .

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$\int_{0}^{\infty} \cos z^2 dz = \sqrt{\frac{\pi}{8}}$

It can be done like this also:

$\int_{0}^{\infty} \frac{\cos x}{\sqrt{x}} dx - i \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}}dx$

$= \ \int_{0}^{\infty} \left ( \frac{\cos x \ - i \sin x }{\sqrt{x}} \right ) dx= \int_{0}^{\infty} \frac{e^{-ix}}{\sqrt{x}} dx= \int_{0}^{\infty}x^{\frac{1}{2} – 1} e^{-ix} dx$

By using the definition of Gamma Function:

$\int_{0}^{\infty} \frac{\cos x}{\sqrt{x}} dx - i \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}}dx = \frac{\Gamma(1/2)}{\sqrt{i}} =  \frac{\sqrt{\pi}}{\sqrt{i}}$

Here, $\sqrt{i} = \left ( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right )^{\frac{1}{2}} = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = \frac{i+1}{\sqrt{2}}$

Hence, $\int_{0}^{\infty} \frac{\cos x}{\sqrt{x}} dx - i \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}}dx = \frac{\sqrt{2\pi}}{i+1} = \sqrt{\frac{\pi}{2}}(1-i)$

So, $\int_{0}^{\infty} \frac{\cos x}{\sqrt{x}} dx =  \sqrt{\frac{\pi}{2}}$

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