help:how to slove this integral

Note by Abhinavyukth Suresh
3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Simply use the Gamma Function. After the substitution ar4=tar^{4}=t, the integral becomes 14a340t14etdt\frac{1}{4a^{\frac{3}{4}}}\int_{0}^{\infty}t^{-\frac{1}{4}}e^{-t}dt. The answer comes out to be Γ(34)4a34\frac{\Gamma\left(\frac{3}{4}\right)}{4a^{\frac{3}{4}}}

Aaghaz Mahajan - 3 months ago

Log in to reply

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

Log in to reply

I have already given the answer tho.

Aaghaz Mahajan - 3 months ago

Log in to reply

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

Log in to reply

@A Former Brilliant Member There is no "proof" neede here. Simply substitute ar4=tar^{4}=t and then the answer follows

Aaghaz Mahajan - 3 months ago

Log in to reply

@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.

Log in to reply

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

Aaghaz Mahajan - 3 months ago

Log in to reply

@Aaghaz Mahajan Maybe his insight into the Gamma Function?

Log in to reply

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

Not saying you're bad or anything.

Log in to reply

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

Abhinavyukth Suresh - 3 months ago

Log in to reply

The proof given by Mr.Aaghaz is correct and mine is same too .Yet using this very idea we can prove this result, Γ(1/n)nζ(1/n)=01exn1dx\frac{\Gamma(1/n)}{n}\zeta(1/n)=\int_0^\infty\frac{1}{e^{x^n}-1}dx, giving us, 1n0x1/n1ex1dx=01exn1dx\frac{1}{n}\int_0^\infty\frac{x^{1/n-1}}{e^x-1}dx=\int_0^\infty\frac{1}{e^{x^n}-1}dxas the result.

Aruna Yumlembam - 3 months ago

Log in to reply

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 - 3 months ago

Log in to reply

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

Aruna Yumlembam - 3 months ago

Log in to reply

@Aruna Yumlembam Sure. Observe that n=1enx = 1ex1\sum_{n=1}^{\infty}e^{-nx}\ =\ \frac{1}{e^{x}-1}

Using this, we have 0xt1ex1dx=0(n=1xt1enx)dx\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 0xaebxdx = a!ba+1\int_{0}^{\infty}x^{a}e^{-bx}dx\ =\ \frac{a!}{b^{a+1}} we will arrive at the answer.

Aaghaz Mahajan - 3 months ago

Log in to reply

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 - 3 months ago

Log in to reply

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..

Log in to reply

Log in to reply

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

Help him, please?

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...