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Sometimes we have to compute integrals that are not easy to calculate so that we need to depend on the method of complex integrals like the residue method. But I became curious about possibility of alternative method of evaluation of integrals other than complex integrals.

For example, do we have an ordinary, 'real' method of integral evaluation method for calculating $$\int_{0}^{\infty}x^{1-\alpha}\cos(\omega x)dx$$ or $$\int_{-\infty}^{\infty}{\cos(ax)\over{b^2-x^2}}dx$$ ? In this question I mean 'real' method in the sense that one does not visit the complex plane to evaluate the integral.

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  • $\begingroup$ Are you sure you meant a minus sign in the denominator of second one instead of plus? :P $\endgroup$ – Pranav Arora May 30 '14 at 13:20
  • $\begingroup$ I know how to evaluate the second one using real method. Is Laplace or Fourier transform OK? $\endgroup$ – Tunk-Fey May 30 '14 at 13:55
  • $\begingroup$ Yes, actually these integrals came up from Fourier transform. But if I have to look up the Laplace transform table, I don't think that's a real solution. Is there a good real method? $\endgroup$ – generic properties May 30 '14 at 14:10
  • $\begingroup$ Following the Pranav's comment, are sure about the minus sign in the denominator? $\endgroup$ – Tunk-Fey May 30 '14 at 14:15
  • $\begingroup$ Well, yes. Actually I originally was interested in $\int_{-\infty}^{\infty}{e^{iax}\over {b^2-(x-ic)^2}}dx$. $\endgroup$ – generic properties May 30 '14 at 14:20
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For $1 <a <2$,

$$ \begin{align} \int_{0}^{\infty} x^{1-a} \cos (wx)\ dx &= w^{a-2} \int_{0}^{\infty} u ^{1-a} \cos (u) \ du \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} \int_{0}^{\infty} \cos (u) \ t^{a-2} e^{-ut} \ dt \ du \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty}t^{a-2} \int_{0}^{\infty} \cos (u)e^{-tu} \ du \ dt \\ & = \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} t^{a-2} \frac{t}{1+t^{2}} \ dt \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} \frac{t^{a-1}}{1+t^{2}} \ dt \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{1}{2} \int_{0}^{\infty} \frac{v^{\frac{a}{2}-1}}{1+v} \ dv \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{1}{2} B \left(\frac{a}{2}, 1- \frac{a}{2} \right) \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{\pi}{2} \csc \left(\frac{\pi a}{2} \right) \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \frac{\pi}{2} \frac{2 \cos \left(\frac{\pi a}{2} \right)}{\sin (\pi a)} \\ &= w^{a-2} \frac{\Gamma(a) \Gamma(1-a) \cos \left(\frac{\pi a}{2} \right)}{\Gamma(a-1)} \\ &= w^{a-2} \ (a-1) \Gamma(1-a) \cos \left(\frac{\pi a}{2} \right) \\ &=- w^{a-2} \ \cos \left(\frac{\pi a}{2} \right) \Gamma(2-a) \end{align}$$

which is the answer given by Wolfram Alpha

If you want to be more rigorous, integrate by parts at the beginning and choose $1+ \sin u$ for the antiderivative of $\cos u$. Then when you switch the order of integration, it's easily justified by Tonelli's theorem.

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  • $\begingroup$ It seems the result is correct but I wonder, does the integral really converge? Let say we take $a=\dfrac13$ and $w=1$, what is the value of integral? $\endgroup$ – Tunk-Fey May 30 '14 at 16:13
  • $\begingroup$ @Tunk-Fey It converges if $a$ is between $1$ and $2$. $\endgroup$ – Random Variable May 30 '14 at 16:17
  • $\begingroup$ Now that makes sense. +1. Anyway, I have the answer for this integral but it involves $i^{\ a-2}$, where I use $i^{\large\ a-2}=e^{\large-\frac{i\pi(2-a)}{2}}$. It does give the same solution as yours. $\endgroup$ – Tunk-Fey May 30 '14 at 16:44
  • $\begingroup$ @Tunk-Fey Thanks. I think I know the approach to which you're referring. As far as I know, the justification for that approach requires contour integration because when you make a complex substitution, you are no longer integrating on the real line. $\endgroup$ – Random Variable May 30 '14 at 17:35
  • $\begingroup$ Indeed you're correct but I use that approach when I was a Physics student and my prof(s) never complaint about that. I think that is the different between physics and math. :D $\endgroup$ – Tunk-Fey May 30 '14 at 17:51
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The first example can be tackled using the expression of the $\Gamma$ function in conjunction with Euler's formula. The second, by differentiating twice under the integral sign, and then solving the resulting functional differential equation, in a manner quite similar to this example.

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    $\begingroup$ By the second one, do you mean by differentiating with respect to $a$? I think it's good idea but it generates divergent integral $\int\cos(ax)dx$... what did I miss? $\endgroup$ – generic properties May 30 '14 at 14:28
  • $\begingroup$ Replace the upper limit with a parameter, and use a technique similar to this or this. $\endgroup$ – Lucian May 30 '14 at 17:45

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