When we calculate the improper integral for $f(x)=\frac{\sin x}{x}$, We think of the following function as $\frac{\exp(iz)}{z}$ whose imaginary part on the real axis is $\frac{\sin x}{x}$. And Contour $C$ will consist of the real axis from $r$ to $R$, the semicircle in the upper half plane from $R$ to $-R$, the real axis from $-R$ to $-r$, and the semicircle in the upper half-plane from $-r$ to $r$. The center of the two semicircles is the origin. We can get the value by using Residue theorem as $\int_{0\leq x} \frac{\sin x}{x} dx = \frac{\pi}{2}$

In fact, this is true even if contour is chosen differently. For example, We choose the semicircle in lower half plane from $-r$ to $r$ even if Cantor does not contain the origin.

I wonder if this can be generalized. I wonder if I am free to choose the appropriate contour(Include singularities or not.) when integrating the function as $g(x) = \frac{1}{x^2-5x+4}$


Yes, you are always free to choose a contour, but the contour you select may be more or less convenient than another one, and you have to be careful with convergence of the resulting pieces. For instance, Jordan's lemma states that

$$ \lim_{R\rightarrow\infty}\int_{C_R}f(z)e^{iaz}dz= 0 $$ if $a>0$, $C_R$ is the upper-half-plane countour, and $\lim_{R\rightarrow\infty}\max_{0\leq\theta\leq\pi}\vert f(Re^{i\theta})\vert = 0$. If $a<0$, you have to use the lower half-plane integral - the upper half-plane one is no longer guaranteed to converge. Similarly, if $a>0$, you have to use the UHP integral - the lower half-plane one might not converge. You can pretty much always flip the "small" contours, but again it tends to be easier to avoid computing residues, so choosing the contour to avoid enclosing poles is usually faster.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.