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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. $\int ^{1} _{0} \frac{1-\cos x}{\frac{x^5}{2}} dx$

out of syllabus now
So does that mean divergence and convergence is no longer in syllabus?
edited
most probably yes. i think we should concentrate only on last 5-6 yrs papers for next yrs pattern
edited by

Here $A),C)$ and $D)$ converges and $B)$ and  diverges.

Because $A),D)$ are definite integral and going to a particular value. So, converges.

Diagram for $C)$ https://www.desmos.com/calculator/ucekshhehy

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

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

So @srestha

if the answer is definite then it converges...

and for divergence ?

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:-

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

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

$I$=$I_{1}-I_{2}$

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