I've been searching the web for a way to prove that $\int^{\infty}_{-\infty}{\sin(x)/x} = \pi$ with complex analysis, because I have a problem of consistency.  I found two, carried in the following link : . But then I wanted to find it by using the fact that :  $\sin(x)/x = \text{Im}(e^{ix}/x)$  To do so, I shifted the integration by $-i$ in the complex plane, showing that the integral on $[-\infty , \infty]$ is equal to the one on $[-\infty -i, \infty -i]$, since the integrand vanish on the vertical borders of the rectangle when we tend to infinity. I did this to get rid of the pole on the contour of integration.  Then, by using the residues theorem on the contour $C_1 : z = y-i, y \in [-R,R]$ and $C_2 : z = Re^{it}-i, t \in [0,\pi]$. Given the integral on $C_2$ vanishes, we have then that :  $\int^{\infty}_{-\infty}{\sin(x)/x} = \text{Im}(\int^{\infty}_{-\infty}{e^{ix}/x}) = \text{Im}(2\pi i\,\text{Res}(e^{iz}/z, 0)) = 2 \pi$  Which gives me the answer with a factor of $2$. I don't understand were did I go wrong ? I think it has something to do with the fact that I take the imaginary part of the integral, but I don't really know... Can someone spot my mistake ?  If needed, I can provide further detail on my calculations.
1632
A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm  dx = \frac \pi 2$$ Well, can anyone prove this without using Residue theory?  I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm  dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.
106185
$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.  I'm trying to understand why and would like some help in working towards a formal proof.
111619
I need help evaluating the following integral $$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$ which should probably be equal to $\frac{\pi}{4}$  Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using integration by parts doesn't lead me to anything pretty.  Update: Not sure if I should post this as a separate question but getting explanation why $\int_{0}^{\infty} \frac{\sin{ax}}{x} = \frac{\pi}{2}$ for a positive integer $a$ could help me solve this question.
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
71517
I'm looking for several ways to prove that $$\int_{0}^{\infty }\sin(x)x^mdx=\cos\left(\frac{\pi m}{2}\right)\Gamma (m+1)$$  for $-2&lt; Re(m)&lt; 0$
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?
49467
$$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$  I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.     My Attempt  This can be rewritten as   $$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$  Now consider   $$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$  Taking the derivative   $$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$  By symmetry we have  $$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$  Using W|A I got   $$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$  By integeration we have   $$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$  Let $x = 2/a$  $$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$  Question  I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
106957
$$\int_0^\pi x \ln(\sin (x))dx  $$   I tried integrating this by parts but I end up getting integral that doesn't converge, which is this $$ \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx$$ So can anyone help me on this one?
24570
Although this question is going to seem completely trivial to anyone with any exposure to path integrals, I'm looking to answer this precisely and haven't been able to find any materials after looking for about 40 minutes, which leads me to believe that it makes sense to ask the question here. In particular I'm looking for an answer wherin any mathematical claims are phrased as precisely as possible, with detailed proofs either provided or referenced. Also my search for a solution has led me to think I'm actually looking for a good explenation of Wick rotation, which I can't really claim to understand in detail. Any good references about this would be very welcome as well.  I'm looking to make sense of the following integral identity:  $$\int_{-\infty}^{\infty} dx \ \exp\left(i\frac{a}{2}x^2+iJx\right)=\left(\frac{2\pi i}{a}\right)^{1/2}\exp\left(\frac{-iJ^2}{2a}\right), \qquad a,J\in\mathbb{R}$$   (and various other sources) say that "This result is valid as an integration in the complex plane as long as a has a positive imaginary part." Clearly the left hand side does not exist in Lebesgue sense for real $a, J$. An answer to the question "" claims:      "it is convergent as a Riemann integral, thanks to some rather delicate cancellations. To make the integral well defined -- equivalently to see how these cancellations occur -- we need to supply some additional information.   Wick rotation provides a way of doing this. You observe that the left hand side is analytic in t   , and that the right hand side is well-defined if Im(t)&lt;0. Then you can define the integral for real t by saying that it's analytic continued from complex t with negative imaginary part."   I want to see the gory details and all known motivation for the validity of this procedure for the kinds of applications where such integrals occur. Suggestions such as "include an $i\epsilon$ in order to make it finite" seem arbitrary. In that case one would have to motivate that prescription from the very start, that is within the modeling procedure that ends up giving that integral expression (which is likely the correct way to approach this). I'm also not sure how to interpret the right hand side, since it involves the square root of an imaginary number, which should involve some choice of branch cut, which I have never seen specified in connection to this formula.
114669
I am trying to solve:  $$\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx$$  and show that it equals $\pi e^{-a}$ for all $a &gt; 0$.  I translate it to its complex equivalent:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$  where C is a half circle with radius R, above the real axis, and $\gamma$ is only the curved (top part) of the half circle.  I've shown that $$\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$ tends to 0 as $R \rightarrow \infty$.  So now:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz$$  By residue theorem,   $$\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f$$  as $ai$ is the pole of $f$ that lies above the real axis.  $$2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}$$  By L'Hospital's rule, this is equivalent to:  $$2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}$$  But how is $\pi i \sin{ai}$ equivalent to $\pi e^{-a}$? Did I make a mistake?
52337
Hi I am trying to find out for what values of the real parameter does the integral $$ I=\int_0^\infty \frac{\sin x}{x^s}dx $$ (a) convergent and (b) absolutely convergent.     I know that the integral is convergent if $s=1$ since $$ \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}. $$ For $s=0$ it is easy to see divergent integral since $\int_0^\infty \sin x\, dx$ is divergent.  However I am stuck on figuring out when it is convergent AND or absolutely convergent.  I know to check for absolute convergence I can determine for an arbitrary series $\sum_{n=0}^\infty a_n$ by considering $$ \sum_{n=0}^\infty |a_n|. $$ If it helps also $$\sin x=\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{(2n+1)!} {x^{2n+1}}$$. Thank you all
27368
Confirm that $\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2$.  The guide book I am using gives the following help:  Consider $\int_{\gamma}z^{-1}e^{iz}dz$, where for $0&lt;s&lt;r&lt;\infty$ the contour of integration is described by $\gamma=[s,r]+\gamma_r+[-r,-s]-\gamma_s$, with $\gamma_r(t)=re^{it}$ and $\gamma_s(t)=se^{it}$ on $[0,\pi]$. Recall Exercise IV$.4.20.$  Exercise IV$.4.20.$ For $r$ with $0&lt;r&lt;\infty$ let $I(r)=\int_{\gamma_r}z^{-1}e^{iz}dz$ with $\gamma_r(t)=re^{it}$ on $[0,\pi]$. Show that $I(r)\rightarrow 0$ as $r\rightarrow \infty$ and also that $I(r)\rightarrow \pi i$ as $r\rightarrow 0$  Using the hint, I know that $\int_{\gamma}z^{-1}e^{iz}dz=0$ for the Cauchy theorem, with which $\int_{[s,r]}z^{-1}e^{iz}dz+\int_{\gamma_r}z^{-1}e^{iz}dz+\int_{[-r,-s]}z^{-1}e^{iz}dz-\int_{\gamma_s}z^{-1}e^{iz}dz=0$, but I do not know what else to do here, could someone help me please? Thank you very much.
4852
I need to prove $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$ for $0&lt;Re(z)&lt;1$. I tried $\cos t=\frac{e^{it}+e^{-it}}{2}$ and to integrate it along the contour from $\epsilon $ to R, then to $Ri$ via a quarter of a circle, then downward to $\epsilon i$, finally back to $\epsilon$ via a small quarter of a circle. However, I find that $e^{-it}t^{z-1} $ does not converge when $R\to \infty$, can anyone give a correction of my integration or offer another method to evaluate the integral?(would be better if using complex contours) thanks in advance.
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?
6312
Another exam problem I'm looking at is to evaluate the following integral. $$ \int_{-\infty}^{\infty} \frac{x\hspace{-0.04 in}\cdot\hspace{-0.04 in}\sin(x)}{x^{\hspace{.02 in}2}+1} dx $$ This is a complex analysis exam, so the solution probably involves contours. $\:$ Since the integrand is an even function, one could potentially simplify by changing one endpoint to $0$. However, I have no idea how to make a contour work since the absolute value of the integrand grows exponentially away from the real axis. What does one need to do to evaluate that integral?
113908
I would like to prove that $$\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$$ diverges without actually evaluating the integral. Is there a convergence test from calculus or real analysis that can show that this integral diverges?  Thanks.  Edit: Someone pointed out that this is a possible duplicate. However, the question put forth as a possible duplicate asks about $\sin(x^2)$, not about $\sin^2(x)$.
103709
$$ \int_{-\infty}^\infty \frac{\sin(x)}{x^2 + 4x + 5}dx$$  What I now is you can use residue theorem, and evaluate this integral  $$ \oint \frac{e^{iz}}{z^2 + 4z + 5}dz$$ and through calculating the residues and using the residue theorem I get that the answer for the above integral is $$\pi e^{-1-2i}$$  However after this I am completely lost.  Any guidance will be much appreciated. I would like to know why I can evaluate the integral as an exponential integral as well.