help:how to slove this integral

#Calculus

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Simply use the Gamma Function. After the substitution $ar^{4}=t$ , the integral becomes $\frac{1}{4a^{\frac{3}{4}}}\int_{0}^{\infty}t^{-\frac{1}{4}}e^{-t}dt$ . The answer comes out to be $\frac{\Gamma\left(\frac{3}{4}\right)}{4a^{\frac{3}{4}}}$

Aaghaz Mahajan - 11 months, 3 weeks ago

You could ask @Aruna Yumlembam - he's an expert on Gamma Functions - he likes them @Abhinavyukth Suresh

A Former Brilliant Member - 11 months, 3 weeks ago

I have already given the answer tho.

Aaghaz Mahajan - 11 months, 3 weeks ago

@Aaghaz Mahajan – He'll probably show the entire proof - see his notes. @Aaghaz Mahajan

A Former Brilliant Member - 11 months, 3 weeks ago

@A Former Brilliant Member – There is no "proof" neede here. Simply substitute $ar^{4}=t$ and then the answer follows

Aaghaz Mahajan - 11 months, 3 weeks ago

@Aaghaz Mahajan – Ok. But I believe he should come. He's good at this stuff. I've read his notes - and gave me a interesting infinite series / function in one of my notes.

A Former Brilliant Member - 11 months, 3 weeks ago

@A Former Brilliant Member – Ok. Although i dont see what else might be needed in the proof.

Aaghaz Mahajan - 11 months, 3 weeks ago

@Aaghaz Mahajan – Maybe his insight into the Gamma Function?

A Former Brilliant Member - 11 months, 3 weeks ago

@Aaghaz Mahajan – Also, you could learn a thing or two from him...

Not saying you're bad or anything.

A Former Brilliant Member - 11 months, 3 weeks ago

so, isn't there any other method than using gamma function to solve this integral?

Abhinavyukth Suresh - 11 months, 3 weeks ago

The proof given by Mr.Aaghaz is correct and mine is same too .Yet using this very idea we can prove this result, $\frac{\Gamma(1/n)}{n}\zeta(1/n)=\int_0^\infty\frac{1}{e^{x^n}-1}dx$ , giving us, $\frac{1}{n}\int_0^\infty\frac{x^{1/n-1}}{e^x-1}dx=\int_0^\infty\frac{1}{e^{x^n}-1}dx$ as the result.

Aruna Yumlembam - 11 months, 3 weeks ago

Yeah we can prove the identity by summing an infinite GP too. Also these types of integrals are known as Bose Einstein Integrals

Aaghaz Mahajan - 11 months, 3 weeks ago

Thanks a lot.But can you please give me a your solution to your problem How is this ??!!

Aruna Yumlembam - 11 months, 3 weeks ago

@Aruna Yumlembam – Sure. Observe that $\sum_{n=1}^{\infty}e^{-nx}\ =\ \frac{1}{e^{x}-1}$

Using this, we have $\int_{0}^{\infty}\frac{x^{t-1}}{e^{x}-1}dx=\int_{0}^{\infty}\left(\sum_{n=1}^{\infty}x^{t-1}e^{-nx}\right)dx$

Swapping the integral and summation , and using the identity $\int_{0}^{\infty}x^{a}e^{-bx}dx\ =\ \frac{a!}{b^{a+1}}$ we will arrive at the answer.

Aaghaz Mahajan - 11 months, 3 weeks ago

Mr. Yajat Shamji,if people provides a solution to any problem you must try and appreciate it and not discourage them.Please don't repeat such acts.

Aruna Yumlembam - 11 months, 3 weeks ago

I'm not. I.. was thinking of you and I read your profile so I thought I could bring you over. After all, you like to contribute, right?

Also, I wasn't discouraging @Aaghaz Mahajan's solution, I..

A Former Brilliant Member - 11 months, 3 weeks ago

@Alak Bhattacharya, @Mahdi Raza, @Zakir Husain, @Gandoff Tan

A Former Brilliant Member - 11 months, 3 weeks ago

@Aruna Yumlembam - @Abhinavyukth Suresh needs your help on solving this integral - needs the Gamma Function and a full proof.

Help him, please?

A Former Brilliant Member - 11 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$