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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
–
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.

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.

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## Comments

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TopNewestSimply 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 $\frac{\Gamma\left(\frac{3}{4}\right)}{4a^{\frac{3}{4}}}$

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

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I have already given the answer tho.

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@Aaghaz Mahajan

He'll probably show the entire proof - see his notes.Log in to reply

$ar^{4}=t$ and then the answer follows

There is no "proof" neede here. Simply substituteLog in to reply

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Not saying you're bad or anything.

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

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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.

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

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

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$\sum_{n=1}^{\infty}e^{-nx}\ =\ \frac{1}{e^{x}-1}$

Sure. Observe thatUsing 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.

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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.

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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..

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

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

Help him, please?

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