Can anyone help me prove the following equivalent found in Peskin &amp; Schroeder on page 27.  $$ \frac{1}{4\pi^2}  \int_{m}^{+\infty} dE \sqrt{E^2-m^2} e^{-iEt} \sim_{t \to \infty} e^{-imt}$$    The integral is divergent, I understand that it has to be understood as distribution but still I'm not sure how to prove it.  I'm looking for a rigorous proof possibly using distribution theory. I am not convinced by the simple change of variable argument.  Edit: This question is indeed a duplicate, however, the answer provided is not as rigorous as I would have liked (using distribution theory) and the given answer I believe is not correct. , the correct equivalent is  $\sim \frac{e^{-imt}}{t \sqrt{t}}$ and not $\sim e^{-imt}$ as claimed in Peskin and Schroeder or $\sim \frac{e^{-imt}}{t}$ as claimed by the answer in the duplicate.
121490
Peskin's QFT textbook   1.page 14  $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}$$  when $x^2\gg t^2$, how do I apply the method of stationary phase to get the book's answer.  2.page 27  $$\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$  where $r&gt;0$  3.page 27  $$\int_{m}^{\infty}\mathrm{d}E \sqrt{E^2-m^2}e^{-iEt}$$  where $m&gt;0$  I'm crazy about these integrals, but the textbook doesn't give the progress.
63713
This  is my problem, true or false?  prove that:   $$\int_{0}^{\infty}\frac{\displaystyle{\sum_{k=1}^{\infty}}k\sin(kx)\,e^{-tk^2}}{\displaystyle{\sum_{k=1}^{\infty}}\cos(kx)\,e^{-tk^2}}dt=\frac{\pi^2({\pi-x})}{8}$$ and $0&lt;x&lt;\dfrac{\pi}{2}$  Thank you.
34849
I'm trying to show that the series $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent.   I'm trying to use a comparison test to do so. To do so, I first can see that $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}$ And I can use a change of variable $k = n+1$, so show $\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty$. Then since $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}$ it must be that $\sum_{k=2}^\infty\frac{1}{k}$ also diverges to infinity. Is this a valid way to show it?
105249
The following is a homework example:  For a nonnegative integer-valued random variable $X$, show that  $\sum^{\infty}_{i=0}{iP(X &gt; i)} = 1/2(E(X²)-E(X)).$  The only way I could think of doing this proof is to rewrite the left side starting with:  $(1).$    $\sum^{\infty}_{i=0}{i} \sum^{\infty}_{j=i+1}{P(X=j}) = 1/2(E(X²)-E(X)).$  and continue from there to get to an expression similar to:  $(2).$    $1/2 * [\sum^{\infty}_{i=0}{i²*P(X=i)} - \sum^{\infty}_{i=0}{i*P(X=i))}] = 1/2(E(X²)-E(X)).$  However I don't know how to continue to transform $(1)$ into $(2)$. Could you give me a hint there? Is the approach even correct?
47463
It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$  Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?  Thanks in advance!
105430
Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$.  I'm not really sure of how I should proceed to show this, and it's pretty un-intuitive as well.  I've managed to manipulate the RHS into  $$\frac{e^{4\beta}}{\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{u}} e^{-(\beta+2u)^2/4u} du$$ which now vaguely resembles a Gaussian. However, substituting $v = \beta + 2u$ in in order to to get something like $e^{-x^2}$ doesn't particularly help. I got $$ \frac{e^{4\beta}}{\sqrt{2\pi}} \int_0^\infty \frac{1}{\sqrt{v - \beta}} e^{-v^2/2(v-\beta)} dv. $$  I'm really unsure of how I should proceed; any help would be appreciated.
18122
I've read the proof for why $\int_0^\infty P(X &gt;x)dx=E[X]$ for nonnegative random variables (located ) and understand its mechanics, but I'm having trouble understanding the intuition behind this formula or why it should be the case at all. Does anyone have any insight on this? I bet I'm missing something obvious.
39730
This integral below $$ I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2} $$ is what I am trying to prove.  Thanks.  We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$.  Next I wrote $$ I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\ \int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx, $$ however only the middle integral is convergent.  I am not sure how to go about solving this problem.  Thank you
48420
How to prove $\int_{\mathbb{R}} e^{-\frac{(x+it)^2}{2}}dx=\sqrt{2\pi}$ for any $t\in \mathbb{R}$? I only obtained the case that $t=0$, $\int_{\mathbb{R}} e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}$.   Thanks.
31221
I want to calculate $$ \int_{0}^{\infty} \frac{\arctan{2t}}{e^{2\pi t}+1} \, {\rm d}t $$ and am wondering if there at all is an analytical expression for the result. I haven't tried much so far, but expressing the denominator as the geometric series and then substituting $z=2\pi t (n+1)$. Equivalently I get $$ \sum_{n=0}^{\infty} \frac{{\rm si}\left(\pi(n+1)\right)}{2\pi (n+1)} $$ where $$ {\rm si}(x) = -\int_{x}^{\infty} \frac{\sin(t)}{t} \, {\rm d}t $$ is the shifted ${\rm Si}(x)$ function. The value is approximately $0.03618247146$, in fact I know from comparison with something else that it is exactly $\frac{\ln \pi -1}{4}$, but there should be some (direct) derivation, or?  Thanks for your input if you have any.  Edit: I tried to follow your instructions, but am now stuck. So I wrote $$ {\cal L}^{-1} \left(\frac{\psi(t)-\psi(1)}{t-1}\right)(s) =\sum_{n=0}^{\infty} \frac{e^{-sn}}{n+1} = e^s \int_s^\infty \frac{{\rm d}x}{e^{x}-1} $$ and taking the Laplace-transform \begin{align} \psi(t) &amp;= \psi(1) + (t-1)\int_0^\infty {\rm d}s \, e^{-s(t-1)} \int_s^\infty \frac{{\rm d}x}{e^{x}-1} \\ &amp;= \psi(1) + (t-1) \int_0^\infty {\rm d}x \int_{0}^{x} {\rm d}s \, \frac{e^{-s(t-1)}}{e^{x}-1} \\ &amp;= \psi(1) + \int_{0}^{\infty} {\rm d}x \, \frac{1-e^{-x(t-1)}}{e^x-1} \tag{1} \\ &amp;= -\frac{1}{t} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-tx}}{e^x-1} \tag{2} \\ &amp;= \dots \\ &amp;=-\sum_{k=0}^{n} \frac{1}{t+k} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-(t+n)x}}{e^x-1} \\ &amp;\stackrel{(\rm{2})}{=} -\frac{1}{2t} + \ln(t) - \int_0^\infty {\rm d}x \, \left\{ \frac{1}{2} - \frac{1}{x} + \frac{1}{e^x-1}\right\} e^{-tx} \tag{Binet 1} \end{align} where $\psi(1)=-\gamma$.  Now we integrate over $t$ from $1$ to $y$ yielding $$ \ln \Gamma(y) \stackrel{({\rm 1})}{=} \psi(1)\,(y-1) + \int_0^\infty {\rm d}x \, \frac{y-1+\frac{e^{-x(y-1)}-1}{x}}{e^x-1} $$ but what now? Substitute?
83094
I am having trouble understanding where the following integral form comes from: $$\int_{-\infty}^{\infty} e^{-a x^2 }e^{-bx}=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}}$$ I see and understand that the value of the improper Gaussian integral is there, and I surmise that a change of variables is necessary when integrating, but I cannot figure out exactly how to prove this answer.
111197
$$\int_{-\infty}^{\infty} \frac{\cos(mx) dx}{e^{-x}+e^x} = \frac{\pi}{e^{m\pi /2}+e^{-m\pi /2}}$$  I need to evaluate the above integral specifically using a rectangluar contour and at some point applying the residue theorem. I should be ok with the residue part, but I'm confused at to how to get this into the example form shown in the book:  $$I =\int_{-\infty}^{\infty} \frac{e^{ax}dx}{1+e^x}=\frac{\pi}{\sin(a\pi)} $$  I tried using the identity $\cos (mx) = \frac{e^{ix}+e^{-ix}}{2}$ but I end up with $\frac{1}{2} \int_{-\infty}^{\infty}\frac{e^{ix}(e^{imx}+e^{-imx})dx}{1+e^{2ix}}$, which almost resembles the example but not quite...
32720
$${\dfrac{1}{\pi^2}}\int_{0}^{\infty}\dfrac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$$   I tried substituting $1/x$ for $x$ but the the only change in the integral is that the $\sqrt{x}$ moves in the numerator from the denominator. I don't understand what to substitute. $\tan{x}$ doesn't seem to work.
16379
Let $X$ and $Z$ be independent, with $X\sim N(0,1)$, and with $\textbf{P}(Z=1)=\textbf{P}(Z=-1)=\frac{1}{2}$. Let $Y=XZ$ (i.e., $Y$ is the product of $X$ and $Z$).   (a) Prove that $Y\sim N(0,1)$.  (b) Prove that $\textbf{P}(|X|=|Y|)=1$.   (c) Prove that $X$ and $Y$ are not independent.   (d) Prove that $\textbf{Cov}(X,Y)=0$.  (e) It is sometimes claimed that if $X$ and $Y$ are normally distributed random variables with $\textbf{Cov}(X,Y)=0$, then $X$ and $Y$ must be independent. Is that claim correct?   My Work: (a) Is really where I am a little stuck. In order to show that $Y\sim N(0,1)$, I want to show that they have the same distribution laws. Thus, I want to show that $\mathcal{L}(Y)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$ given that $\mathcal{L}(X)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$.   However, I don't know how to go about that. Another approach I have considered is to show that for any Borel measurable function $f:\mathbb{R}\to\mathbb{R}$,$\space$ $\textbf{E}(f(X))=\textbf{E}(f(Y))$ for which each is well defined. Which approach do you think is better and How should I start. Any help will be appreciated.
62670
Does anyone have  a proof for $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.$$ I tried to get it from contour integrating $$\frac{e^{iz^2}-1}{z^2},$$ but failed. Thanks.
68926
I want to find the following integral: $$\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$$ I have been trying to apply simple substitutions like $u=e^x$, to simplify the integral without any luck. I simplified it to  $$\int_{0}^{\infty} \frac{1}{x(e^x+1)}-\frac{1}{x(e^{3x}+1)}dx$$  which seems to have increased my problems.  Wolfram Alpha gives the answer to be 0.5493... and the graph looks like a Gaussian distribution.  Somehow, I feel that this integral is very non-trivial due to the 'troublesome' $x$ in the denominator.
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.