How Do You Solve This Complex Integral Involving Cosine and a Square Root?

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Discussion Overview

The discussion revolves around solving a complex integral involving cosine and a square root, specifically the integral \(\frac{2}{\pi}\int \frac{cos(ux)}{\sqrt{x}} dx\) from zero to infinity. The scope includes mathematical reasoning and exploration of different approaches to evaluate the integral.

Discussion Character

  • Mathematical reasoning, Exploratory

Main Points Raised

  • One participant requests help with the integral and provides the specific form of the integral they are trying to solve.
  • Another participant suggests using a substitution \(y^2 = x\) based on knowledge of Gaussian integrals.
  • A participant reports their attempt with the substitution, yielding an answer of \(\frac{\sqrt{2}}{\sqrt{u\pi}}\), but questions the correctness of the expected answer \(\frac{1}{\sqrt{u}}\).
  • Another participant challenges the correctness of \(\frac{1}{\sqrt{u}}\) and notes that both Maple and Wolfram Alpha support the earlier result of \(\frac{\sqrt{2}}{\sqrt{u\pi}}\).
  • The initial poster acknowledges a mistake and expresses gratitude for the assistance, indicating a positive reception of the tools mentioned.

Areas of Agreement / Disagreement

The discussion shows some disagreement regarding the correct evaluation of the integral, with multiple viewpoints on the expected result. There is no consensus reached on the final answer.

Contextual Notes

Participants reference computational tools like Maple and Wolfram Alpha, but there are unresolved questions about the assumptions and steps taken in the calculations.

nemanja
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Hi,i would be gratefull if anyone could help me with this problem.

[tex]\frac{2}{\pi}[/tex][tex]\int \frac{cos(ux)}{\sqrt{x}} dx[/tex]
x goes from zero to infinity.

thanks in advance.
 
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Assuming you know a thing or two about Gaussian integrals, you could try substituting [itex]y^2 = x[/itex]...
 
yes I tried that and I got[tex]\frac{\sqrt{2}}{\sqrt{u\pi}}[/tex]
but correct is [tex]\frac{1}{\sqrt{u}}[/tex] .
any other way of solving?
 
Are you sure 1/√u is correct? Maple and wolfram alpha agree with your answer.
 
Ok my mistake,thank you all.Maple rules :smile:
 

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