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数学代写|复分析代写Complex analysis代考|MATH3979 Spiking it with trig

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数学代写|复分析代写Complex analysis代考|Spiking it with trig

The integrals
$$\int_{-\infty}^{\infty} \frac{\sin (2 x)}{x^4+81} d x, \quad \int_{-\infty}^{\infty} \frac{\cos (5 x)}{\left(x^2+1\right)\left(x^2+36\right)} d x$$
appear, at first sight, to be very different from those we studied in Section 10.2. Certainly the theory of (real) partial fractions will not help us to break up these integrands into simpler pieces-it applies only to pure rational functions. And if we do the obvious thing, converting them into complex functions merely by replacing $x$ by $z$ thus:
$$\int \frac{\sin (2 z)}{z^4+81} d z, \quad \int \frac{\cos (5 z)}{\left(z^2+1\right)\left(z^2+36\right)} d z$$
it doesn’t help a great deal: in particular because the complex sine and cosine are more complicated functions than their real counterparts. (For instance, the useful inequalities $|\sin x| \leq 1,|\cos x| \leq 1$, which simplify so many estimates in real calculus, do not carry over to the complex case: the complex sine and cosine are both unbounded.)

However, if instead we recall Euler’s formula and perceive $\cos \theta$ and $\sin \theta$ as the real and imaginary parts of $e^{i \theta}$, something much more encouraging begins to unfold. For one thing, $e^{i a z}$ does not have any poles or zeros, so the modified integrals
$$\int \frac{e^{2 i z}}{z^4+81} d z, \quad \int \frac{e^{5 i z}}{\left(z^2+1\right)\left(z^2+36\right)} d z$$
experience no more (and no fewer) residues than were already present in their associated rational functions. More encouraging (and less obvious) is the fact that Lemma 10.2.2, concerning the disappearance of integrals of certain types of function along the infinite semicircle, is still perfectly valid for this sort of function also:

10.3.1 Lemma For any bottom-heavy rational $f(z)=\frac{p(z)}{q(z)}$ and any positive real constant $a$, the integral $\int_{S_K} e^{i a z} f(z) d z$ around the semicircular arc $S_K$ tends to zero as $K \rightarrow \infty$.

Proof Recall that when we write out $f(z)$ in full detail:
$$f(z)=\frac{a_0+a_1 z+a_2 z^2+\cdots+a_n z^n}{b_0+b_1 z+b_2 z^2+\cdots \cdots+b_m z^m}$$
we can reshape it as
$$f(z)=\frac{1}{z^2} r(z)$$
where $r(z)$ tends to a limit as $|z| \rightarrow \infty$, and is consequently bounded: we can find a constant $M$ such that $|r(z)| \leq M$ provided that $|z|$ is big enough.

So, what effect does the extra factor $e^{i a z}$ have on the integral? Expressing $z$ as $x+i y$ we see that
$$\left|e^{i a z}\right|=\left|e^{i a x-a y}\right|=\left|e^{i a x}\right|\left|e^{-a y}\right|=e^{-a y}$$

数学代写|复分析代写Complex analysis代考|Some special case techniques

The unit circle and the so-called infinite semicircle are only two out of an enormous range of simple closed contours that have proved useful in evaluating real integrals via complex techniques: indeed, there are so many ‘special cases’ that all we can reasonably do is to illustrate a handful of them. The common thread is, once you have decided on a complex function that on the real axis will give you the real function that you want, you need to select a contour along whose edges the integrals will be either very small or directly computable or closely related to the ‘real’ integral that is being sought and, in particular, a contour that avoids the function’s singularities. We start with a variation on the ‘bottom-heavy rationals’ of Sections $10.2$ and $10.3$ where, as it turns out, the previous choice of contour is still powerful enough to deliver results.

数学代写|复分析代写Complex analysis代考|Spiking it with trig

$$\int_{-\infty}^{\infty} \frac{\sin (2 x)}{x^4+81} d x, \quad \int_{-\infty}^{\infty} \frac{\cos (5 x)}{\left(x^2+1\right)\left(x^2+36\right)} d x$$

$$\int \frac{\sin (2 z)}{z^4+81} d z, \quad \int \frac{\cos (5 z)}{\left(z^2+1\right)\left(z^2+36\right)} d z$$

$$\int \frac{e^{2 i z}}{z^4+81} d z, \quad \int \frac{e^{5 i z}}{\left(z^2+1\right)\left(z^2+36\right)} d z$$

10.3.1 引理对于任何重底有理数 $f(z)=\frac{p(z)}{q(z)}$ 和任何正实常数 $a$, 积分 $\int_{S_K} e^{i a z} f(z) d z$ 围荛半圆弧 $S_K$ 趋于零 $K \rightarrow \infty$.

$$f(z)=\frac{a_0+a_1 z+a_2 z^2+\cdots+a_n z^n}{b_0+b_1 z+b_2 z^2+\cdots \cdots+b_m z^m}$$

$$f(z)=\frac{1}{z^2} r(z)$$

$$\left|e^{i a z}\right|=\left|e^{i a x-a y}\right|=\left|e^{i a x}\right|\left|e^{-a y}\right|=e^{-a y}$$

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