I want to know how to solve this using contour integration:  $$\int_0^{\infty} \frac{\sin(x)}{\sqrt{x}}dx.$$  So I let the integral become:  $$\oint_c \frac{\sin(z)}{\sqrt{z}}dz$$ where c is a "half doughnut" shape avoiding the singularity at z = 0 and extending into the upper half of the complex plane towards infinity.   $$\oint_c = \int_{up} + \int_{-R}^{- \epsilon} + \int_{low} + \int_ {\epsilon}^R = 0$$  (Because no singularities are actually contained within the contour.)  By a bound argument, the $\int_{up}$ contributes nothing to the integral. Therefore:  $$\lim{R \to \infty}, {\epsilon \to 0}$$  $$- \int_{low} = \int_{- \infty}^{\infty} $$  Where $\int_{low}$ is the integral over the bump going over the point z = 0. So can I use the Cauchy Integral theorem to say  $$\int_{low} = \pi i\, \text{Res} \left( \frac{\sin x}{ \sqrt{x}}, 0 \right)$$   Because there is no residue for this function, which would imply the integral is zero, which I know it is actually $\sqrt \frac{ \pi}{2}$.
70652
Numerically it seems to be true that  $$ \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. $$  Any ideas how to prove this?
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?
89496
How do you evaluate the following integral? Here we take $a&gt;1$.  $$\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{\frac{3}{2}}}.$$  I know I have to use the Residue Theorem, however, I am stuck on which contour to use, and also how to find the pole of the function. Any hints are greatly appreciated.
120960
Prove $$\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} \text{d}x = 8\pi$$   I tried using a modified version of the integrand and an origin-indented semicircular contour on the positive real half of the complex plane.  Despite my best efforts, I am unable to retrieve the desired integral, as the negative factor from the negative imaginary axis is preventing me from being able to simplify the integrals.  On top of that, the residue doesn't come anywhere close to the exact result.  Are there better methods that I am missing?
120781
I ran across this integral I get no where with. Can someone suggest a method of attack?.  $$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$  I tried series, imaginary parts, and so forth, but have made no progress.    Thanks very much.
48343
The following integral has been on my mind for a while $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x \tag{$\dagger$}$$    Let us indicate the integrand as $f(x)=\frac{\sin(x)}{e^x-1}$. The following are a couple of observations.   The integrand can be extended by continuity in $0$ since $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\frac{\sin(x)}{x}}{\frac{e^x-1}{x}}=1$$ This is the original reason I started playing around with this integral. Mathematica yields the result $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x=\frac{\pi}{2}\textrm{Coth}(\pi)-\frac 12 \approx 1.076674047$$ which, following numerical evidence, seems correct. Complex analysis may be useful here, since the integrand is a holomorphic function on $\mathbb C$. I tried writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, but I was not able to find an appropriate integration contour to solve the problem. The $-1$ in the denominator breaks the simmetry of the expression. This made most of my substitutions useless.   Can this integral be evaluated correctly, preferrably through complex analytic methods?
37377
Recently I found a claim saying that  $$  \int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2  $$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.   Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$ \int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x $$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the , but there are some subtle differences.
26610
I'm trying to compute $\displaystyle \int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}dx$, $(0&lt;a&lt;1)$ Let $f$ denote the integrand.  I'm using the rectangular contour given by the following curves: $c_1: z(t) = R+it, t \in [0, 2\pi]$ $c_2: z(t) = -t+2\pi i, t \in [-R, R]$ $c_3: z(t) = -R + i (2\pi - t), t \in [0, 2\pi]$ $c_4: z(t) = t, t \in [-R, R]$    There is one singularity within the contour, at $z = \pi i$. Expanding out the denominator as a power series shows that it's a simple pole, and allows us to evaluate the residue as $\displaystyle \lim_{z \rightarrow \pi i} f(z)(z-\pi i) = - e^{a \pi i}$    This is computed by expanding $1+e^z$ as a Taylor series around $\pi i$. The first coefficient will be 0, and the second will be $-1$. The rest will have orders of $(z - \pi i)$ greater than 1, and will thus vanish when we take the limit.  So the integral over the entire contour is $- 2\pi i e^{a \pi i}$    An easy enough estimate on the $c_1$ shows that the integral vanishes as $R \rightarrow \infty$. With a variable change, c_3 is the same as c_1 and also vanishes. $c_4$ becomes the integral we want when we take a limit. $c_2$ becomes $c_4$ with a constant:    \begin{align*} \int_{c_2} f(z)dz &amp;= \int_{-R}^{R} \frac{e^{-at}e^{a 2\pi i}}{1+e^{-t}e^{2\pi i}}dt \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{-at}}{1+e^{-t}}dt \\ &amp;=e^{a 2 \pi i}\int_{R}^{-R} - \frac{e^{au}}{1+e^{u}}du \ \ \ (u = -t, du = -dt)   \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{au}}{1+e^{u}}du \\ &amp;= e^{a 2 \pi i} I(R) \end{align*}    Where $I(R)$ is the line integral over $c_4$. Putting it all together and taking the limit gives us $\displaystyle \lim_{R \rightarrow \infty} I(R) = \frac{- 2\pi i e^{a \pi i}}{(1 + e^{a 2 \pi i}) }$    But this can't be the value of the integral, because it's a real-valued function integrated over $R$. I can't figure out where I'm going wrong. Note that I've avoided posting all the details of my solution since this is from a current problem set for a class on complex analysis.
64675
$$\int_{0}^{\infty }\sin(x^{2})dx$$  I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.  I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions (without complex analysis)?
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.
5470
I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.  Here is a version of a rational log integral rarely encountered.  $\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$  I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.  My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?.  Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?.
59712
In my book I have a theorem that goes something like the following     Let $P(x)$ be $Q(x)$ polynomials such that $\deg(Q) \geq \deg(P) + 2$.   Then \begin{align*}  \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}  e^{iax} \,\mathrm{d}x  = 2 \pi i \frac{a}{|a|} \sum_{k=1}^{m} \mathrm{Res}\left[  \frac{P(x)}{Q(x)} , z_k \right] \end{align*} where $a$ is a real   constant and $z_1,\,\ldots\,,z_m$ are the singularities to $P(x)/Q(x)$ in   the upper halfplane if $a&gt;0$ and the lower halfplane if $a&lt;0$.   I am having a bit of problems understanding this. I know the reason why we have to switch contour is that $e^{iz}$ blows up. But I really can not see why or how it blows up.   Take the canonical example for why this theorem is useful  $$   J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1} $$    Now the clue here is to show that the integral along the curve tends to zero, and then use the residue theorem. To show that the integral tends to zero I did this \begin{align*}  \left| \int_{C_1} \frac{e^{iz}}{1+z^2} \,\mathrm{d}z\right|  \leq \int_{C_1} \left| \frac{e^{iz}}{1+z^2} \right| \,\mathrm{d}z   \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{\left|1+z^2\right|} \right)       \int_{C_1} |e^{iz}| \,\mathrm{d}z \end{align*} where the $ML$-inequality was used in the last inequality  On the circle with radius $R$ we have $|e^{iz}|=R$, and we can use the inequality $|a+b|\leq|a|-|b|$ to simplify further \begin{align*}  \left|\int_{D_R} \frac{e^{iz}}{1+z^2}\,\mathrm{d}z \right|  \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{|z|^2-1}\right)  \int_{D_R} R \,\mathrm{d}z  \leq \pi \frac{R}{R^2-1} \end{align*} which tends to zero as $R \to \infty$ as wanted. but if one instead had $e^{-iz}$ then the theorem states that one has to use the lower half plane. But I do not see where my calculations err if one persists in using the contour in the upper half plane? What goes wrong, and why does it go wrong? $|e^{-iz}|$ should still be $R$ on the semi-circle.   How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane?
53974
Using residues, try the contour below with $R \rightarrow  \infty$ and  $$\lim_{R \rightarrow  \infty }  \int_0^R  \frac{1}{1+r^n}  dr \rightarrow \int_0^\infty  \frac{1}{1+x^n}  dx$$    I've attempted the residue summation, but my sum did not converge.
88631
Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .  Here is what i have tried :  I found out that for all $x&gt;0$ , $x-\frac{x^3}{6} &lt; \sin x &lt;x $  . So $1-\frac{x^2}{6}&lt;\frac{ \sin x}{x}&lt;1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}&lt;\int_{0}^{M}|\frac{ \sin x}{x}|&lt; M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved .  Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .
59248
How do I go about evaluating the following by contour integration?  $$ \int^1_0 \frac{dx}{(x^{2} - x^{3})^{1/3}} $$  The question does not fit in the standard form of : $\int^{2\pi}_0$ or $\int^\infty_{-\infty}$  etc.  I tried substituting $ x = cos(\theta) $ but got stuck.
29140
Could anyone help me decide what contour to use to evaluate this integral?  $$\int_0^\infty \frac{\sqrt{t}}{1+t^2} dt$$  So we have simple poles at $i$,$-i$. Why does using a quarter of a circle in the upper right quadrant not work -is it a problem having $i$ on the contour of integration?  My second idea was to let $$\sqrt{t}=e^{\frac{1}{2}\operatorname{log}t}$$  And defining the branch of the logarithm to be the negative imaginary axis. Then using an upper semi-circular contour. With a hole at 0, which will tend to the point 0 as the radius of the semi circle tends to infinity. Then we'll divide by two to get the integral from $0$ to $\infty$. Does this work, as this does not seem to be an even function we're integrating.  I apolygise for the poor explanation of my contours, it's quite difficult without pen and paper!  EDIT: Ok, so using an upper semi circle, with a semicircular hole about $0$. let us call the contour $\gamma$, we have:  $$\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt =2\pi i\Big(\frac{e^{\frac{1}{2}\operatorname{log}t}}{t+i}\Big)\Big|_{t=i}=\frac{e^{\pi/4}}{2i}$$  But by letting on the positive real axis $z=x+iy$ implies $log(z)=log(x)+0$ and on the negative axis $log(z)=log(x)+\pi$ Therefore if we tend the large arc's radius to infinity and the small arc about the origin's radius to zero we find the integral along the arc tends to zero. So  $$\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt=\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx+\int_0^\infty \frac{\sqrt{x}e^{\pi/2}}{1+x^2} dx=(1+e^{\pi/2})\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx$$  $$\Rightarrow \int_0^\infty \frac{\sqrt{x}}{1+x^2} =\frac{e^{\pi/4}}{2i(1+e^{\pi/2})}$$  This is wrong but can anyone point out what mistake I have made here?
70240
Compute the Riemann integral  $$\int_0^{\frac\pi2}\frac{\mathrm d \theta}{\sqrt{\sin \theta}}$$  It seems very difficult, I don't know how to go ahead.  Thank you very much for your help!