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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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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

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