Please Some one provide me the Solution Of This Problem .

And

Please Some one provide me the Solution Of This Problem (2) .

And

Please Some one provide me the Solution Of This Problem (3)

Please Some one provide me the Solution Of This Problem .

And

Please Some one provide me the Solution Of This Problem (2) .

And

Please Some one provide me the Solution Of This Problem (3)

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TopNewestHey guys please also solve the 3rd Problem. – Rishabh Deep Singh · 1 year ago

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I have also added a new question Please check it Also If anyone can help me solving it. – Rishabh Deep Singh · 1 year ago

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– Rishabh Cool · 1 year ago

Have you got the second question??Log in to reply

– Rishabh Deep Singh · 1 year ago

Can you post the solution of It.Log in to reply

– Rishabh Cool · 1 year ago

It is a bit long..... Apply integration by parts to evaluate the integral and then apply limit....Log in to reply

– Rishabh Deep Singh · 1 year ago

Oka Thanks.Log in to reply

– Rishabh Cool · 1 year ago

Welcome.. :-)Log in to reply

It is very easy. Just a matter of 5-6 steps. Use Ramanujan's Master Theorem. I'll provide a solution after 3 April. – Aditya Kumar · 1 year ago

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– Rishabh Deep Singh · 1 year ago

Please send Us solution Now. I need it. Please!Log in to reply

– Aditya Kumar · 1 year ago

My parents aren't allowing me to use PC. Shall I send you a pic through whatsapp?Log in to reply

– Rishabh Deep Singh · 1 year ago

Yes sure 8558060484Log in to reply

– Aditya Kumar · 1 year ago

I had missed a term that's why I got it as infinity.Log in to reply

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– Rishabh Deep Singh · 1 year ago

This question was in my Definite Integration Module . Yes it converges.Log in to reply

I feel its somewhat related to Riemann Sums.But fortunately we have a cool guy in our community: @Rishabh Cool who will post an awesome solution to that problem :) – Nihar Mahajan · 1 year ago

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@Rishabh Deep Singh can help by telling what he has done to crack the problem.... DO you have a hint Nihar?? – Rishabh Cool · 1 year ago

Unfortunately I had tried that problem many times..... But I'm not able to find from where to start ..... Probably that integration sign is doing most of the harm... There might be a way to get rid of that integration sign.... ProbablyLog in to reply

@Pi Han Goh @Ishan Singh @Aditya Kumar – Nihar Mahajan · 1 year ago

Time to tag seniors:Log in to reply

First question.Step 1) Notice that the integrand is an even function, so \( \displaystyle \int_{-\infty}^{\infty} \cdots = 2 \int_{0}^{\infty} \cdots \).

Step 2) Break the integral into two parts: \(\displaystyle 2 \int_{0}^{\infty} \cdots = 2 \int_{0}^{1} \cdots + 2 \int_{1}^{\infty} \cdots\).

Step 3) For the latter integral, try a substitution of \(y = 1/x\).

Step 4) Combine the 2 integrals into 1. Have you tried GP?

Step 5) Profit.

Second question.Do you know how to differentiate a Beta function? Differentiate through an integral

Step 1) \(\displaystyle B(a,b) = \dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \int_0^{\pi /2} (\sin x)^a (\cos x)^b \, dx \).

Step 2) Differentiiate both sides with respect to \(b\).

Step 3) What is the derivative of a Beta function?

Step 4) Profit. – Pi Han Goh · 1 year ago

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– Rishabh Deep Singh · 1 year ago

I don't know the differentiation of Beta function.Log in to reply

Correct me if I am wrong. – Harsh Shrivastava · 1 year ago

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– Pi Han Goh · 1 year ago

Limits are indeed from 0 to pi/2. So it's possible.Log in to reply

– Rishabh Cool · 1 year ago

2nd one can also be solved by first evaluating the integral using IBP and then applying limit.... :-)Log in to reply

– Pi Han Goh · 1 year ago

POST! POST! POST! POST! POST! POST! POST! POST!Log in to reply

– Rishabh Cool · 1 year ago

\[=\lim_{a\to \frac{\pi}{2}^-} (\ln (\cos a)^{\sin a-1})\] how to evaluate this limit ??Log in to reply

– Pi Han Goh · 1 year ago

Why are you posting the actual solution here? Go post it as solution that problem itself.Log in to reply

– Rishabh Cool · 1 year ago

Sorry... I just wanted to know how to calculate that limit..Log in to reply

– Pi Han Goh · 1 year ago

Haha don't need to apologize. Have you tried l hospital rule?Log in to reply

– Rishabh Cool · 1 year ago

Oh thanks... Done L'hopital did the job... This was the only point where I struggled in the question but now its done .... Thanks :-)Log in to reply