# Is there always an alternative way to compute integral other than complex integral?

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.

• Are you sure you meant a minus sign in the denominator of second one instead of plus? :P – Pranav Arora May 30 '14 at 13:20
• I know how to evaluate the second one using real method. Is Laplace or Fourier transform OK? – Tunk-Fey May 30 '14 at 13:55
• 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? – generic properties May 30 '14 at 14:10
• Following the Pranav's comment, are sure about the minus sign in the denominator? – Tunk-Fey May 30 '14 at 14:15
• Well, yes. Actually I originally was interested in $\int_{-\infty}^{\infty}{e^{iax}\over {b^2-(x-ic)^2}}dx$. – generic properties May 30 '14 at 14:20

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.

• 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? – Tunk-Fey May 30 '14 at 16:13
• @Tunk-Fey It converges if $a$ is between $1$ and $2$. – Random Variable May 30 '14 at 16:17
• 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. – Tunk-Fey May 30 '14 at 16:44
• @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. – Random Variable May 30 '14 at 17:35
• 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 – Tunk-Fey May 30 '14 at 17:51

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.

• 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? – generic properties May 30 '14 at 14:28
• Replace the upper limit with a parameter, and use a technique similar to this or this. – Lucian May 30 '14 at 17:45