Here is a persuasive advertisement for ghostwriting services for the topic of contour integration and substitution methods:
**Unlock the Secrets of Complex Integrals with Our Expert Ghostwriting Services**
Are you struggling to evaluate complex integrals like ∫e^(iz^2)/z dz? Do you need help mastering contour integration and substitution methods? Our team of expert ghostwriters is here to help!
**Our Services:**
* Customized solutions for complex integrals, including contour integration and substitution methods
* Expertly crafted explanations and step-by-step guides for easy understanding
* High-quality, plagiarism-free content tailored to your specific needs
* Fast turnaround times to meet your deadlines
**Why Choose Us:**
* Our writers are experts in mathematics, with advanced degrees and extensive experience in contour integration and substitution methods
* We guarantee accurate and reliable solutions, backed by rigorous research and attention to detail
* Our services are confidential and secure, ensuring your academic integrity is protected
**Get Ahead of the Curve:**
Don’t let complex integrals hold you back from achieving academic success. Our ghostwriting services will give you the edge you need to excel in your mathematics course. Order now and receive:
* A comprehensive solution to ∫e^(iz^2)/z dz and other complex integrals
* A deeper understanding of contour integration and substitution methods
* More free time to focus on other important aspects of your academic life
**Order Now and Unlock the Secrets of Complex Integrals!**
Contact us today to learn more about our ghostwriting services and take the first step towards mastering complex integrals.
\[\int_{\gamma_{1}}\frac{e^{iz^{2}}}{z}dz=\int_{0}^{\pi/2}\frac{e^{i(Re^{i\theta})^{ 2}}}{Re^{i\theta}}Rie^{i\theta}d\theta=i\int_{0}^{\pi/2}e^{i(Re^{i\theta})^{2}}d\theta.\]
Integral along\(-i[r,R]\): On \(-i[r,R]\), we have that \(z=0+iy\). Therefore, \(dz=idy\) and
\[\frac{e^{iz^{2}}}{z}=\frac{e^{i(iy)^{2}}}{iy}=\frac{e^{-iy^{2}}}{iy}.\]
Thus
\[\int\limits_{-i[r,R]}\frac{e^{iz^{2}}}{z}dz=-\int\limits_{i[r,R]}\frac{e^{iz^{ 2}}}{z}dz=-\int_{r}^{R}\frac{e^{-iy^{2}}}{iy}idy=-\int_{r}^{R}\frac{e^{-}iy^{2}} {y}dy.\]
Integral along \(-\gamma_{2}\): On \(\gamma_{2}\), we have that \(z=\gamma(\theta)=re^{i\theta}\). This implies that \(dz=rie^{i\theta}d\theta\) and \(e^{iz^{2}}=e^{i(re^{i\theta})^{2}}\). Therefore we have
\[\int\limits_{-\gamma_{2}}\frac{e^{iz^{2}}}{z}dz=-\int\limits_{\gamma_{2}} \frac{e^{iz^{2}}}{z}dz=-\int_{0}^{\pi/2}\frac{e^{i(re^{i\theta})^{2}}}{re^{i \theta}}rie^{i\theta}d\theta=-i\int_{0}^{\pi/2}e^{i(re^{i\theta})^{2}}d\theta.\]
Thus rewriting equation (3.4), we have that
\[\int_{r}^{R}\frac{e^{ix^{2}}}{x}dx+i\int_{0}^{\pi/2}e^{i(Re^{i\theta})^{2}}d \theta-\int_{r}^{R}\frac{e^{-iy^{2}}}{y}dy-i\int_{0}^{\pi/2}e^{i(re^{i\theta}) ^{2}}d\theta=0.\]
The first and third above integrals can be combined as
\[\int_{r}^{R}\frac{e^{ix^{2}}-e^{-ix^{2}}}{x}dx.\]
Recalling the identity \(\sin(x^{2})=\frac{e^{ix^{2}}-e^{-ix^{2}}}{2i}\), the integral becomes
\[2i\int_{r}^{R}\frac{\sin(x^{2})}{x}dx.\]
Since our problem is to evaluate \(\int_{0}^{\infty}\frac{\sin(x^{2})}{x}dx\), we must take \(R\rightarrow\infty\) and \(r\to 0\) and get estimates for the 2nd and 4th integrals.
Claim #1: The second integral in (3.4) vanishes as \(R\rightarrow\infty\).
Proof: Observe that
\[\left|i\int_{0}^{\pi/2}e^{i(Re^{i\theta})^{2}}d\theta\right|\leq\int_{0}^{\pi /2}\left|e^{i(Re^{i\theta})^{2}}\right|d\theta. \tag{3.5} \]However,\]However,
\[\left|e^{i(Re^{i\theta})^{2}}\right|=\left|e^{iR^{2}e^{2i\theta}}\right|=\left|e^ {i[R^{2}(\cos(2\theta)+i\sin(2\theta))]}\right|=e^{-R^{2}\sin(2\theta)}.\]
(Note: recall that \(|e^{iz}|=|e^{i(x+iy)}|=|e^{ix}\cdot e^{-y}|=|e^{ix}|e^{-y}=e^{-y}\)). Going back to (3.5), we get
\[\left|i\int_{0}^{\pi/2}e^{i(Re^{i\theta})^{2}}d\theta\right|\leq\int_{0}^{\pi/ 2}e^{-R^{2}\sin(2\theta)}d\theta=2\int_{0}^{\pi/4}e^{-R^{2}\sin(2\theta)}d\theta.\]
Think of \(\sin(2\theta)\). The graph looks like Figure 3.37.
When \(\theta\in\left[0,\frac{\pi}{4}\right]\), we have that \(\sin(2\theta)\geq\frac{4\theta}{\pi}\).
Note: one can also see \(\sin(2\theta)\geq\frac{4\theta}{\pi}\) by setting up a function \(h(\theta)=\sin(2\theta)-\frac{4\theta}{\pi}\) and observing that \(h(0)=0\) and \(h(\frac{\pi}{4})=0\) while \(h^{\prime\prime}(0)<0\) for \(0<\theta<\frac{\pi}{4}\), implying that \(\sin(2\theta)\geq\frac{4\theta}{\pi}\).
Continuing with (3.5), we see that
\[2\int_{0}^{\pi/4}e^{-R^{2}\sin(2\theta)}d\theta<2\int_{0}^{\pi/4}e^{-R^{2} \frac{4\theta}{\pi}}d\theta.\]
We can integrate this last integral. In fact,
\[2\int_{0}^{\pi/4}e^{-\frac{4R^{2}}{\pi}\theta}d\theta=-2\frac{\pi}{4R^{2}}e^{ -\frac{4R^{2}}{\pi}\theta}\Big{|}_{0}^{\pi/4}=-\frac{\pi}{2R^{2}}[e^{-R^{2}}- 1]=\frac{\pi}{2R^{2}}[1-e^{-R^{2}}].\]
Now as \(R\to\infty\), the second integral in (3.4) will therefore vanish.
**Claim #2:** The fourth integral is equal to \(i\frac{\pi}{2}\) as \(r\to 0\).
Proof: We use the continuity of \(e^{iz^{2}}\) at \(z=0\). When \(z=0\), \(e^{iz^{2}}=1\). Given any \(\epsilon>0\) there must exist a \(r>0\) such that
\[|e^{iz^{2}}-1|<\epsilon\]whenever \(|z| \[\left|i\int_{0}^{\pi/2}e^{i(re^{i\theta})^{2}}d\theta-i\pi/2\right|=\left|i\int_{0 }^{\pi/2}\left(e^{i(re^{i\theta})^{2}}-1\right)d\theta\right|<\epsilon\cdot \frac{\pi}{2}.\] Thus the th integral approaches \(i\frac{\pi}{2}\) as \(r\to 0\). Going back to equation (3.4), combining the st and rd integrals and using Claim #1 and Claim #2, we have \[\int_{0}^{\infty}\frac{e^{ix^{2}}-e^{-ix^{2}}}{x}dx-i\frac{\pi}{2}=2i\int_{0}^{ \infty}\frac{\sin(x^{2})}{x}dx-i\frac{\pi}{2}=0.\] Thus \[\int_{0}^{\infty}\frac{\sin(x^{2})}{x}dx=\frac{\pi}{4}\] as desired. ### Exercises 1. Evaluate the following integrals: 1. \(\int_{\gamma}e^{z}\,dz\) where \(\gamma\) is the perimeter of the unit square 2. \(\int_{\gamma}\frac{1}{z^{2}}\,dz\) where \(\gamma\) is the unit circle 3. \(\int_{\gamma}\frac{1}{z}\,dz\) where \(\gamma\) is the circle \(3+e^{i\theta}\), \(0\leq\theta\leq 2\pi\) 4. \(\int_{\gamma}z^{2}dz\) where \(\gamma\) is the segment joining \(1+i\) to \(2\) 3. Does Cauchy’s theorem hold separately for real and imaginary parts of \(f\)? If so, prove that it does; if not, give a counterexample. 4. Let \(\gamma\) be a piecewise smooth closed curve. Show that \[\int_{\gamma}z^{m}dz=0,\ \ \ \ \ m=0,1,2,\ldots\] 5. Use Green’s theorem (complex form) to show that \[\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-p}=\left\{\begin{array}{ll}0&\ \ \ \ \mbox{if $p$ is outside $\gamma$}\\ 1&\ \ \ \ \mbox{if $p$ is inside $\gamma$.}\end{array}\right.\]Let \(D\) be an arbitrary region and \(R=\{x+iy:a\leq x\leq b,\ \ c\leq y\leq d\}\) be a rectangle. Show that if \(R\subseteq D\) and \(f\) is differentiable in \(D\), then \[\int_{\partial R}f=0,\] where \(\partial R=[z_{1},z_{2}]+[z_{2},z_{3}]+[z_{3},z_{4}]+[z_{4},z_{1}]\). 7. Let \(f(z)\) be holomorphic on a simple connected region \(A\), except possibly not holomorphic at \(z_{0}\in A\). Suppose, however, that \(f\) is bounded in absolute value near \(z_{0}\). Show that for any simple closed curve \(\gamma\) containing \(z_{0}\), \[\int_{\gamma}f(z)\,dz=0.\] 8. Let \(f\) be holomorphic on \(\mathbb{C}\) (such functions are called entire functions). Evaluate \[\int_{0}^{2\pi}f(z_{0}+re^{i\theta})e^{i\theta}d\theta.\] 9. Let \(f\) be a continuous function in the disc \(|z-z_{0}|
2. For what simple closed curves \(\gamma\) does the following hold: \[\int_{\gamma}\frac{dz}{z^{2}+z+1}=0.\]
发表回复