Thread to share Good Questions from Calculus which requires deep thinking


Aise to sarre integral zero hone chahiye na but yeh ni hai


@Yash_Srivastava2 Very well done bro, I was also searching for some integral which would oppose the conclusion arrived from improper substitution

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Got it why to check that , u r right @Rohan_Shinde1 bro

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Jisse tum numerator me x ko put kr sako

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Without the inverse how will we be able to substitute the x in integrand? Moreover the inverse must also be differentiable within confined limits so that you replace the differential dx with appropriate substituting differential dt in our case.


Put x-1 =t
Then use Taylor series.

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@Yash_Srivastava2 @Azimuddin_Sheikh

Precisely this is what we were missing. So we in fact have to find Cauchy Principle Value of the integral given.


But I'm not getting how to evaluate the series.
\sum_{r=1}{\infty} \frac{1}{2^{r} r^2} plz help
I remember I did this for limit 0 to 1 and got \zeta( 2)


Hey @Rohan_Shinde1 are these questions solvable by methods in jee syllabus or some out of jee course.
Asking because solving out of syllabus question won't help when 3 months is left for JEE😀


Bro in this ques i m converting limit to 0 to 1
Then getting infinite HP with x-1/2 in numerator but unable to solve further
I m doing correctly na??


The series is perfectly correct. In fact it converges to \frac {\pi^2}{12}-\frac {\ln^2 2}{2}

Well done.


Don't worry some of them, approx. 2 to 3 questions will require a little higher advanced methods but others should do fine


Bro as far as I know
That is the polylogrithmic function ,right?
So \displaystyle P_4(1)=\sum_{k=1}^{\infty} \left(\frac{1}{k^4} \right)

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How you found the sum of that series...????

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Stupendous... Marvelous.....

Guessed it right, they are called the \color{red}{\text{Polylogarithms}}

About P_4(1)=\zeta(4)=\frac {\pi^4}{90}
So that too is correct


What about this bro ques 2 @Rohan_Shinde1


Bro are these questions given to you by somebody or these are your extraworks.
Too :open_mouth:heartbeating questions
This is the link very useful


@Yash Shrivastava I didn't quite understand what you meant so I would pen down my solution. If you still get doubts feel free to ask.

\displaystyle I=\int_1^{\infty} \frac {\{x\}-\frac 12}{x}dx
\displaystyle =\lim_{n\to\infty} \left(\int_1^n dx -\int_1^n \frac {\lfloor x\rfloor +\frac 12}{x}dx\right)
\displaystyle =\lim_{n\to\infty} \left( (n-1)-\frac 12\left(\ln\left(\frac {2^3}{1^3}\cdot \frac{3^5}{2^5}\cdot \frac {4^7}{3^7}.....\frac{n^{2n+1}}{(n-1)^{2n+1}}\right)\right) \right)
\displaystyle =-1+\lim_{n\to\infty} \left(n-\frac 12\ln\left(\frac {n^{2n+1}}{(n!)^2}\right) \right)

Now just Stirling's approximation for factorials to reach the answer which is -1+\ln (\sqrt {2\pi})


I already have read quite a much about it. Coming to the questions, some of them are original while some are from internet or some random book.


@Rohan_Shinde1 i was doing via another method but was getting same result as in second last step but was stuck there

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Those worrying about only JEE scope should then skip \color{red}{\text{Q. 6, Q.7, Q.8}} ..... I guess Q.7 is also doable

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