$$\int_{0}^{+\infty}\frac{\sin x}{x}dx$$     My start:  Zero is a singular point, Let's define $g(z):=(e^{iz})\big/z$  $$\int_{\Gamma}\frac{e^{iz}}{z}dz=\\ \int_{-R}^{-r}\frac{e^{ix}}{x}dx+\int_{C_r^+}\frac{e^{iz}}{z}dz+\int_{r}^{R}\frac{e^{ix}}{x}dx+\int_{C_R^+}\frac{e^{iz}}{z}dz$$  $$=\int_{r}^{R}\frac{e^{ix}-e^{-ix}}{x}dx+\int_{C_r^+}\frac{e^{iz}}{z}dz+\int_{C_R^+}\frac{e^{iz}}{z}dz$$  from here I'm stuck the final answer should be $\pi /2$  $$(*)\lim\limits_{r\to 0}\int_{C_r^+}\frac{e^{iz}}{z}dz$$  $$(**)\lim\limits_{R\to 0}\int_{C_r^+}\frac{e^{iz}}{z}dz$$
123246
How can I calculate $$\int_0^\infty\frac{\sin x}xdx$$ by integration of a suitable function along the following paths:  where $R$ and $\varepsilon$ are the radius of the shown outer and inner semicircle, respectively. This seems to be quite interesting. How would one do that?
29631
There are three parts of the question, the first two which I proved.   $a.$ Proving $f(z)$ is entire analytic.  $b.$ $f(z+1) = f(z)$ and $f(z+i) = e^{\pi}e^{-2\pi iz}f(z)$  $c.$ Inside the unit square in the 1st quadrant, prove that $f(z)$ has a unique $z_0$ such that $f(z_0)=0$  I need some help regarding the $c.$ part of the question. I don't understand how to apply the argument principle here (which is a hint). It is also given that $f(z)$ does not vanish on the boundary of the square.  What I could do:  $$f(0) = f(1) = 1+2\sum_{n=1}^{\infty}e^{-\pi n^2}$$  and   $$f(i) = f(1+i) = e^{\pi}\big(1+2\sum_{n=1}^{\infty}e^{-\pi n^2}\big)$$  Using the hint given below by a user: I end up with $$\dfrac{1}{2\pi i}log f(z)$$ and the integral evaluates to -1/2 and 1/2 on the two vertical sides and 0 on the horizontal lines. Resulting in the closed integral to 0. Any ideas where I might be going wrong?  Any idea how to proceed?
87202
Using a rectangular contour in the complex plane, bypassing the poles at $z=0$ and $z=i$, i got $$\int_{-\infty}^{+\infty}\frac{\sinh(ax)e^{ikx}}{\sinh(\pi x)}dx - e^{-k}\int_{-\infty}^{+\infty}\frac{\sinh[a(x+i)]e^{ikx}}{\sinh(\pi x)}dx = \sin(a)e^{-k}$$  and now?
35050
I have to calculate $$\int_0^\infty\frac{\log x dx}{x^2-1},$$  and the hint is to integrate $\frac{\log z}{z^2-1}$ over the boundary of the domain $$\{z\,:\,r&lt;|z|&lt;R,\,\Re (z)&gt;0,\,\Im (z)&gt;0\}.$$  I don't understand. The boundary of this domain has a pole of the integrand in it, doesn't it? Doesn't it make this method useless?
116076
I need to prove that if $f$ and $g$ are analytic in $D_r(z_0)$ and $g$ has a simple zero at $z_0$, then $$\mathrm{Res}[f/g,z_0]=\frac{f(z_0)}{g'(z_0)}$$   When $f(z_0)\neq 0$ and since $1/g$ has a simple pole at $z_0$ $$\mathrm{Res}[f/g,z_0]=\lim_{z\to z_0}(z-z_0)\frac{f(z)}{g(z)}=\frac{f(z_0)}{g'(z_0)}$$ But what do I need to do when $f(z_0)=0$? Any help will be appreciated thanks
113167
Let's consider the following integral $$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx. \tag{1}$$  I would like to compute $(1)$ using residue theory. Let's consider a complex function $$f(z) = \frac{\cos2z}{(z^2+4)^2} = \frac{\cos2x}{(z-2i)^2(z+2i)^2}.$$  Of course $\text{Im}(-2i) &lt; 0$ thus I am to calculate residue only in the point $z_0 = 2i$.  Noting that $z_0$ is a double pole we have $$R = \text{res}_{z_0}f(z) = \lim_{z \to 2i} \frac{d}{dz} \bigg((z-2i)^2 \frac{\cos(2z)}{(z-2i)^2(z+2i)^2} \bigg).$$  After some calculations we get $$R = \frac{i \big(-5 + 3 e^8 \big)}{64 e^4}.$$ That implies $$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx = 2 \pi i \frac{i \big(-5 + 3 e^8 \big)}{64 e^4} = -\frac{(-5 + 3 e^8) \pi}{32 e^4}.$$  According to  $(1)$ is equal to $$\frac{5 \pi}{16 e^4}.$$  What am I doing wrong?
74912
I would like to show $\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$.  Now, of course, the integrand does not converge uniformly to $0$ on $\theta\in [0, \pi/2]$, since it has value $1$ at $\theta =0$ for all $r\in \mathbb{R}$.   If $F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$, we can find the $j$th derivative $F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$, but I don't see how this is helping.  The function is strictly decreasing on $[0,\pi/2]$, since $\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$, which is strictly negative on $(0,\pi/2)$.  Any ideas?
111255
$$\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^2(x^2+16)}$$  My attempt:  We are integrating over the real axis, which is the real part of a set of complex numbers, so $$=\int_{-\infty}^{\infty} \frac{dz}{(z^2+1)^2(z^2+16)}$$  $$=\int_{-\infty}^{\infty} \frac{dz}{((z+i)(z-i))^2(z+4i)(z-4i)}$$  But I notice that all of the singularities are imaginary! Should this give me pause? What should I do now?
64295
$$\int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .  Attempt  $[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right?  Though the answer given is $\dfrac{17}{2}$.
97606
Let $0&lt;\alpha&lt;2.$ Looking for a proof for the following: $$\int_0^{+\infty}\frac{\sin(x)}{x^\alpha}dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}.$$ Any ideas?
45819
$$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx$$   Some rearranging eventually gives $$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx = \frac{-i}{2}\int_{-\infty}^{\infty} \frac{xe^{2ix}}{9+x^{2}}$$ Consider $f(z) = \frac{ze^{2iz}}{9+z^{2}}$ and the contour $\gamma$ of the semicircle laying in the upper half of the plane:   Let $\gamma_{R}$ denote the circular part with radius $R$ and $\gamma_{L}$ denote the part lying on the real axis with length $2R$.   Computing the residue at the only pole, $z = 3i$, we have that $$\oint_{\gamma}f(z) \, dz = \frac{i\pi}{e^{6}} $$ On the other hand, \begin{align*} \oint_{\gamma}f(z) \, dz &amp;= \oint_{\gamma_{R}}f(z) \, dz + \oint_{\gamma_{L}} f(z) \, dz \\ &amp;= \oint_{\gamma_{R}} f(z) \, dz + \int_{-R}^{R} \frac{xe^{2ix}}{9+x^{2}} \, dx \end{align*} We may evaluate the first integral in the usual way by parameterizing the contour and taking $z = Re^{i\theta}$. \begin{align*} \oint_{\gamma_{R}} f(z) \, dz = \int_{0}^{\pi} \, \frac{R^{2}e^{2i\theta}e^{2i\cos\theta}e^{-2R\sin\theta}}{(9+R^{2}e^{2i\theta})} d\theta \end{align*}  I'm aiming to show that this integral goes to $0$ as $R$ goes to infinity. This gives the desired result as the factor of $\frac{-i}{2}$ is all that is missing according to WolframAlpha. I'm not sure how to finish it though.
66834
The question is as follows (note that $z = x + yi$):     Does $\lim_\limits{z \to 0} e^{-1/z^4}$ exist?   I've shown that      $\lim_\limits{x \to 0} e^{-1/x^4} = 0$   is equal to      $\lim_\limits{y \to 0} e^{-1/(iy)^4} = 0$   To prove both parts, but I'm not sure this is how it is done. I've been away from class for surgery, so all I have to work off of is what I've read online and in the textbook.   The problem is, it seems that with problems such as this one:     $\lim_\limits{z \to 2} \frac{z^2 + 3}{iz} = \frac{-7i}{2}$   All you have to do is substitute in 2 directly. I can't seem to figure out how to approach these problems, so hopefully all I have explained makes sense. Thanks for any explanations!
90359
Let $\displaystyle f$ be an entire function such that $$\lim_{|z|\rightarrow \infty} |f(z)| = \infty .$$ Then,    $f(\frac {1}{z})$ has an essential singularity at 0. $f$ cannot be a polynomial. $f$ has finitely many zeros. $f(\frac {1}{z})$ has a pole at 0.   Please suggest which of the options seem correct.   I am thinking that $f$ can be a polynomial and so option (2) does not hold.  Further, if $f(z) = \sin z $ then it has infinitely many zeros... which rules out (3) while for $f(z) = z$ indicates that it has a simple pole at $0$ and option (4) seems correct.
33764
I need to show that  $$ \int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3} $$  I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from.  Using a change of variable $ x \mapsto 2x $ :  $$ \int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi $$  Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain  $$ \int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$  But I am now at a loss as to how to make $x^4$ appear at the denominator.  Any ideas appreciated.  Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
57939
$$\lim_{n \to \infty }n\int_{0}^{1}\frac{x^{n}}{1+x+x^{n}}dx$$  I factorize $x^n$ then I tried with $u=1+x^{1-n}$ but I didn't get too far. Also I tried to get something from $0&lt;x&lt;1$. I know that if $x$ is from $(0,1)$ then $x^n$ tends to $0$ as $n \to \infty$.
46433
I'm asked to show  $\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}$ when $t,a &gt; 0$. I tried using integration by parts integrating $\frac{x}{x^2+t^2}$. But it seems that $\frac{x\sin ax}{x^2+t^2}$ has no antiderivative. What should I try?
77573
I need to calculate    $$I_0=\int_{-\infty}^\infty\frac{e^{ix}}{x}dx=\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{ix}}{x}dx=\lim_{R\rightarrow 0}(\lim_{\varepsilon\rightarrow 0}(\int_{-R}^\varepsilon\frac{e^{ix}}{x}dx+\int_{-\varepsilon}^R\frac{e^{ix}}{x}dx))=\lim_{R\rightarrow 0}(\lim_{\varepsilon\rightarrow 0}I(R,\varepsilon))$$   I have tried to close the contour of the complex integral on the complex plain. Let $C$ be the semicircle with radius R above the real axis and $\gamma$ be the semicircle under the real axis with radius $\varepsilon$, both centered at $O$. Let $L$ be the closed contour contained $C,[-R,-\varepsilon],\gamma,[\varepsilon,R]$. Denote the $+$ direction by the $+$ direction of $L$. Then we have:  $$\int_{L^+}\frac{e^{iz}}{z}dz=I(R,\varepsilon)+\int_{C^+}\frac{e^{iz}}{z}dz+\int_{\gamma^+}\frac{e^{iz}}{z}dz$$  By Cauchy Integral Formula we have $\int_{L^+}\frac{e^{iz}}{z}dz=2\pi i$ so  $$I(R,\varepsilon)=2\pi i-\int_{C^+}\frac{e^{iz}}{z}dz-\int_{\gamma^+}\frac{e^{iz}}{z}dz$$  Take $\varepsilon\rightarrow 0$ we have:  $$I(R,0)=\pi i-\int_{C^+}\frac{e^{iz}}{z}dz=\pi i-\int_0^{\pi}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}iRe^{i\phi}d\phi$$  $$=\pi i-i\int_0^\pi\exp(iR(\cos\phi+i\sin\phi))d\phi=\pi i-i\int_0^\pi e^{iR\cos\phi}e^{-R\sin\phi}d\phi$$  Now I must take the limit $R\rightarrow\infty$. I evaluate:  $$\left|\int_0^\pi e^{iR\cos\phi}e^{-R\sin\phi}d\phi\right|\le\int_0^\pi e^{-R\sin\phi}d\phi$$  Is there anything wrong in my argument so far? How can I proceed further?  EDIT1: Looking at it for quite long I think I can write  $$\int_0^\pi e^{-R\sin\phi}d\phi=2\int_0^{\pi/2} e^{-R\sin\phi}d\phi\le 2\int_0^{\pi/2} e^{-R\frac{\phi}{2}}d\phi=\frac{4}{R}(1-e^{-R\frac{\pi}{4}})\le\frac{4}{R}$$  So in the end take $R\rightarrow\infty$ we get $I_0=i\pi$. Is it true?