忘记什么时候搞来的积分了，有些好像是积分竞赛的，应该很多都不是我自己做的，只是原来记录的

$\begin{aligned} &\dfrac{z}{1-z}=\sum_{n=1}^{\infty}z^n\qquad(|z|<1)\\
&左右两边积分可得;\ln(1-z)=-\sum_{n=1}^\infty\dfrac{z^n}{n}.\\
&代入z=re^{i\theta},\dfrac{re^{i\theta}}{1-re^{i\theta}}=\sum_{n=1}^{\infty}r^n(\cos n\theta+i\sin n\theta)\\
&左边=\dfrac{re^{i\theta}}{1-re^{i\theta}}=\dfrac{r\cos\theta+ir\sin\theta}{(1-r\cos\theta)-ir\sin\theta}=\dfrac{(r\cos\theta+ir\sin\theta)[(1-r\cos\theta)+ir\sin\theta]}{(1-r\cos\theta)^2+r^2\sin^2\theta}\\
&\qquad=\dfrac{r\cos\theta-r^2}{1-2r\cos\theta+r^2}+i\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}\\
&因此\dfrac{\cos\theta-r}{1-2r\cos\theta+r^2}=\sum_{n=1}^{\infty}r^{n-1}\cos n\theta,\quad \dfrac{\sin\theta}{1-2r\cos\theta+r^2}=\sum_{n=1}^{\infty}r^{n-1}\sin n\theta.\\
&左右对r积分:\quad\dfrac12\ln(1-2r\cos\theta+r^2)=-\sum_{n=1}^{\infty}r^n\dfrac{\cos n\theta}{n}\cdots①\\
&\qquad\qquad\qquad\quad \arctan\dfrac{r\sin\theta}{1-r\cos\theta}=\sum_{n=1}^{\infty}r^n\dfrac{\sin n\theta}{n}\cdots②\\
&将①式中r换为-r得：\\
&\qquad\qquad r\cos\theta-r^2\dfrac{\cos2\theta}{2}+r^3\dfrac{\cos3\theta}{3}-\cdots=\dfrac12\ln(1+2r\cos\theta+r^2)\\
&将②式中的r换为-r，\theta换为-\theta得：\\
&\qquad\qquad r\sin\theta-r^2\dfrac{\sin2\theta}{2}+r^3\dfrac{\sin3\theta}{3}-\cdots=\arctan\dfrac{r\sin\theta}{1+r\cos\theta} \end{aligned}$

$ \begin{aligned} \displaystyle&\int\dfrac{\sin \theta}{r^2-2r\cos\theta+1}\text{d}r\\
=&\int\dfrac{\sin\theta}{(r-\cos\theta)^2+\sin^2\theta}\text{d}(r-\cos\theta)\\
=&\arctan\dfrac{r-\cos\theta}{\sin \theta} \end{aligned} $

$ \begin{aligned} \left(\arctan\dfrac{r\sin\theta}{1-r\cos\theta}\right)'=&\dfrac{1}{\left(\dfrac{r\sin\theta}{1-r\cos\theta}\right)^2+1}\cdot\dfrac{\sin\theta(1-r\cos\theta)+\cos\theta\cdot r\sin\theta}{(1-r\cos\theta)^2}\\
=&\dfrac{\sin\theta}{r^2\sin^2\theta+1-2r\cos\theta+r^2\cos^2\theta}\\
=&\dfrac{\sin\theta}{1-2r\cos\theta+r^2} \end{aligned} $

$ \begin{aligned} &\dfrac{\sin\theta}{1-2r\cos\theta+r^2}=\dfrac{\sin\theta}{1-2r\cos\theta+r^2\cos^2\theta+r^2\sin^2\theta}\\
=&\dfrac{\sin\theta}{(1-r\cos\theta)^2+r^2\sin^2\theta}=\dfrac{\dfrac{\sin\theta}{(1-r\cos\theta)^2}}{1+\left(\dfrac{r\sin\theta}{1-r\cos\theta}\right)^2} \end{aligned} $

$\displaystyle\int_0^r\dfrac{\dfrac{\sin\theta}{(1-r\cos\theta)^2}}{1+\left(\dfrac{r\sin\theta}{1-r\cos\theta}\right)^2}\text{d}r=\int_0^r\dfrac{1}{1+\left(\dfrac{r\sin\theta}{1-r\cos\theta}\right)^2}\text{d}(\dfrac{r\sin\theta}{1-r\cos\theta})\\=\arctan\dfrac{r\sin\theta}{1-r\cos\theta}$

$\displaystyle\int_0^1\dfrac{\arctan x}{x\sqrt{1-x^2}}\text{d}x=\int_0^{\frac{\pi}{4}}\dfrac{x}{\tan x}$

$\displaystyle\lim_{x\to0^+}\ln(1-x)\ln(\tan x)=\lim_{x\to0^+}\ln(1-x)\ln(\sin x)=\lim_{x\to0^+}-x\ln(\sin x)=\lim_{x\to0^+}-\dfrac{\ln(\sin x)}{\frac1x}=\lim_{x\to0^+}\dfrac{\cos x}{\sin x}\cdot x^2=0$

  • $\cos\theta+\dfrac{\cos2\theta}{2}+\dfrac{\cos3\theta}{3}+\dfrac{\cos4\theta}{4}+\cdots$

直接引用昨天的结论：

$r\cos\theta-r^2\dfrac{\cos2\theta}{2}+r^3\dfrac{\cos3\theta}{3}-\cdots=\dfrac12\ln(1+2r\cos\theta+r^2)\cdots①;$
把 $r=-1$ 带入 ① 式，得：

$-\cos\theta-\dfrac{\cos2\theta}{2}-\dfrac{\cos3\theta}{3}-\cdots=\dfrac12\ln(2-2r\cos\theta)=\ln(2\sin\dfrac{\theta}{2})$

$\cos\theta+\dfrac{\cos2\theta}{2}+\dfrac{\cos3\theta}{3}+\cdots=-\ln(2\sin\dfrac{\theta}{2})$

  • $\sin\theta+\dfrac{\sin2\theta}{2}+\dfrac{\sin3\theta}{3}+\dfrac{\sin4\theta}{4}+\cdots$

直接引用昨天的结论：

$r\sin\theta-r^2\dfrac{\sin2\theta}{2}+r^3\dfrac{\sin3\theta}{3}-\cdots=\arctan\dfrac{r\sin\theta}{1+r\cos\theta}\cdots②$
把 $r=-1$ 带入 ② 式中，得：
$-\sin\theta-\dfrac{\sin2\theta}{2}-\dfrac{\sin3\theta}{3}-\cdots=\arctan\dfrac{-\sin\theta}{1-\cos\theta}=\arctan\dfrac{-\cos\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}}$

$\sin\theta+\dfrac{\sin2\theta}{2}+\dfrac{\sin3\theta}{3}+\cdots=\arctan\dfrac{\cos\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}}=\arctan\cot\dfrac{\theta}{2}=\dfrac{\pi}{2}-\arctan\tan\dfrac{\theta}{2}=\dfrac{1}{2}(\pi-\theta)$

  • $\cos\theta+\dfrac{\cos3\theta}{3}+\dfrac{\cos5\theta}{5}+\dfrac{\cos7\theta}{7}+\cdots$

原式 $\displaystyle=\sum_{n=1}^{\infty}\dfrac{\cos(2n-1)\theta}{2n-1}$

使用昨天的中间结论：
$\displaystyle\dfrac{\sin\theta}{1-2r\cos\theta+r^2}=\sum_{n=1}^{\infty}r^{n-1}\sin n\theta$
把 $r=-1$ 和 $r=1$ 带入上式得：

$\dfrac{\sin\theta}{2+2\cos\theta}=\sin\theta-\sin2\theta+\sin3\theta-\cdots$

$\dfrac{\sin\theta}{2-2\cos\theta}=\sin\theta+\sin2\theta+\sin3\theta+\cdots$

对 $\theta$ 求导，得：
$-\sin\theta-\sin3\theta-\sin5\theta-\sin7\theta-\cdots=-\dfrac{1}{2}\left(\dfrac{\sin\theta}{2+2\cos\theta}+\dfrac{\sin\theta}{2-2\cos\theta}\right)=-\dfrac{1}{4}(\dfrac{\sin\dfrac{\theta}{2}}{\cos\dfrac{\theta}{2}}+\dfrac{\cos\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}})=-\dfrac{1}{4\sin\dfrac{\theta}{2}\cdot\cos\dfrac{\theta}{2}}$

左右两边对 $\theta$ 积分：
$\cos\theta+\dfrac{\cos3\theta}{3}+\dfrac{\cos5\theta}{5}+\dfrac{\cos7\theta}{7}+\cdots=\dfrac{1}{2}\ln\cot\dfrac{\theta}{2}$

  • $\sin\theta+\dfrac{\sin3\theta}{3}+\dfrac{\sin5\theta}{5}+\dfrac{\sin7\theta}{7}+\cdots$

原式 $\displaystyle=\sum_{n=1}^{\infty}\dfrac{\sin(2n-1)\theta}{2n-1}$

使用昨天的中间结论：
$\displaystyle\dfrac{\cos\theta-r}{1-2r\cos\theta+r^2}=\sum_{n=1}^{\infty}r^{n-1}\cos n\theta$
把 $r=-1$ 和 $r=1$ 带入上式得：

$\displaystyle\dfrac{\cos\theta+1}{2+2\cos\theta}=\cos\theta-\cos2\theta+\cos3\theta-\cdots=\dfrac12$

$\displaystyle\dfrac{\cos\theta-1}{2-2\cos\theta}=\cos\theta+\cos2\theta+\cos3\theta+\cdots=-\dfrac12$

左右两边相加得：
$\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta+\cdots=\dfrac{\cos\theta+1}{2+2\cos\theta}+\dfrac{\cos\theta-1}{2-2\cos\theta}=0$

左右两边对 $\theta$ 积分：
$\sin\theta+\dfrac{\sin3\theta}{3}+\dfrac{\sin5\theta}{5}+\dfrac{\sin7\theta}{7}+\cdots=C$

  1. 当 $\theta=0$ 时，$\sin\theta+\dfrac{\sin3\theta}{3}+\dfrac{\sin5\theta}{5}+\dfrac{\sin7\theta}{7}+\cdots=0$
  2. 当 $\theta=\dfrac{\pi}{2}$ 时，$\displaystyle\sin\theta+\dfrac{\sin3\theta}{3}+\dfrac{\sin5\theta}{5}+\dfrac{\sin7\theta}{7}+\cdots=\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{2n-1}=\dfrac{\pi}{4}$

综上 $\sin\theta+\dfrac{\sin3\theta}{3}+\dfrac{\sin5\theta}{5}+\dfrac{\sin7\theta}{7}+\cdots=\dfrac{\pi}{4}\text{sgn}\theta$

  • $\sin\theta-\dfrac{\sin3\theta}{3^2}+\dfrac{\sin5\theta}{5^2}-\dfrac{\sin7\theta}{7^2}+\cdots=\dfrac{\pi}{4}\theta\qquad-\dfrac{\pi}{2}\leq\theta\leq\dfrac{\pi}{2}$

左右两边对 $\theta$ 求导，得：
$\cos\theta-\dfrac{\cos3\theta}{3}+\dfrac{\cos5\theta}{5}-\dfrac{\cos7\theta}{7}+\cdots$

左右两边对 $\theta$ 求导，得：
$\displaystyle-\sin\theta+\sin3\theta-\sin5\theta+\cdots=\sum_{n=1}^{\infty}(-1)^n\sin(2n-1)\theta$

$\sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$

  • $\displaystyle\cos\theta-\dfrac{\cos5\theta}{5}+\dfrac{\cos7\theta}{7}-\dfrac{\cos11\theta}{11}+\cdots=\dfrac{\pi}{2\sqrt3}=\dfrac{2}{\sqrt3}\sum_{n=1}^{\infty}\sin\dfrac{(2k-1)\pi}{3}\dfrac{\cos(2k-1)\theta}{2k-1},\qquad-\dfrac{\pi}{3}\leq\theta\leq\dfrac{\pi}{3}$

$\displaystyle\cos\theta-\dfrac{\cos5\theta}{5}+\dfrac{\cos7\theta}{7}-\dfrac{\cos11\theta}{11}+\cdots=\dfrac{2}{\sqrt3}\sum_{n=1}^{\infty}\sin\dfrac{(2k-1)\pi}{3}\dfrac{\cos(2k-1)\theta}{2k-1}$

$\displaystyle\int_0^1\left(\left\lfloor\dfrac{2}{x}\right\rfloor-2\left\lfloor\dfrac{1}{x}\right\rfloor\right)\text{d}x=2\ln2-1$

$\displaystyle\int_0^1\left(\left\lfloor\dfrac{k}{x}\right\rfloor-k\left\lfloor\dfrac{1}{x}\right\rfloor\right)\text{d}x=k\ln k-k\left(\dfrac12+\dfrac13+\cdots+\dfrac1k\right)$

$\displaystyle I=\int_0^{\arctan k}\dfrac{x}{\sin^2x+k\sin x\cos x+2\cos^2x}\text{d}x$

设 $x=\arctan k-t$ .且注意到：
$$f(x)=\sin^2x+k\sin x\cos x+2\cos^2x=\frac12[k\sin2x+\cos2x]+\dfrac32$$ 则有：
$$f(\arctan k-x)=\dfrac12[k\sin2(\arctan k-x)+\cos2(\arctan k-x)]+\dfrac32=f(x)$$ $$I=\int_0^{\arctan k}\dfrac{\arctan k-t}{f(t)}\text{d}t=\arctan k\int_0^{\arctan k}\dfrac{1}{f(x)}\text{d}x-I$$ 由于：
$$\int\dfrac{1}{f(x)}\text{d}x=\int\dfrac{1}{k\sin2x+\cos2x+3}\text{d}x=\dfrac{2\arctan\left(\dfrac{k+2\tan x}{\sqrt{8-k^2}}\right)}{\sqrt{8-k^2}}+C$$ 故
$$I=\dfrac{\arctan k\left(\arctan\dfrac{3k}{\sqrt{8-k^2}}-\arctan \dfrac{k}{\sqrt{8-k^2}}\right)}{\sqrt{8-k^2}}$$

$\displaystyle\int_0^1\dfrac{\arctan x}{1+x^2}\ln\left(\dfrac{1+x^2}{1+x}\right)\text{d}x$

设 $x=\tan t$ .则：
$$I=\int_0^{\frac\pi4}t\ln\left(\dfrac{1}{\cos t(\sin t+\cos t)}\right)\text{d}t$$
设 $f(x)=\cos x(\sin x+\cos x)$, 则有： $$f(\dfrac{\pi}{4}-x)=f(x)$$ 设 $t=\frac\pi4-u$. 那么有
$$I=\int_0^{\frac\pi4}t\ln\dfrac{1}{f(t)}\text{d}t=\int_0^{\frac\pi4}(\dfrac{\pi}{4}-u)\dfrac{1}{f(\frac\pi4-u)}\text{d}u=\dfrac{\pi}{4}\int_0^{\frac\pi4}\ln\dfrac{1}{f(u)}\text{d}u-I$$ 只要计算： $$f=\int_0^{\frac\pi4}\ln(\cos^2t+\sin t\cos t)\text{d}t=\int_0^{\frac\pi4}\ln(\cos x)+\int_0^{\frac\pi4}\ln(\sqrt{2}\cos(x-\frac\pi4))\text{d}x$$ 利用卡特兰常数的一个熟悉积分 $$\int_0^{\frac\pi4}\ln(\cos x)\text{d}x=\dfrac{G}{2}-\dfrac{\pi}{4}\ln2$$ 带入得 $$I=\dfrac{3\pi^2\ln2}{64}-\dfrac{G}{8}$$

$\displaystyle\int_0^{\pi}\ln(\sin x+\sqrt{1+\sin^2x})\text{d}x$

$\begin{aligned} I=&\int_0^\pi\ln(\sin x+\sqrt{1+\sin^2x})\text{d}x\\
=&2\int_0^{\frac\pi2}\ln(\sin x+\sqrt{1+\sin^2x})\text{d}x\\
=&2\int_0^{\frac\pi2}\ln(1+\sqrt{1+\frac{1}{\sin^2x}})\text{d}x-\pi\ln2\\
I(a)=&2\int_0^{\frac\pi2}\ln(1+\sqrt{1+\frac{a}{\sin^2x}})\text{d}x\\
I'(a)=&\int_0^{\frac\pi2}\dfrac{\frac1{\sin^2x}}{\sqrt{1+\frac{a}{\sin^2x}}+1+\frac{a}{\sin^2x}}\text{d}x\quad t=\cot x\\
=&\int_0^{+\infty}\dfrac{\text{d}t}{\sqrt{(1+a)+at^2}+(1+a)+at^2}\\
&\qquad\quad t=\sqrt{\dfrac{1+a}{a}}\tan u\\
=&\dfrac{\pi-2\arctan \frac1{\sqrt a }}{2a}\\
=&\dfrac{\arctan \sqrt a }{a}\\
I=&I(1)-\pi\ln2=\int_0^1\dfrac{\arctan \sqrt a }{a}\text{d}a+\pi\ln2-\pi\ln2\\
=&\int_0^1\dfrac{\arctan\sqrt{a}}{a}\text{d}a=2\int_0^1\dfrac{\arctan t}{t}\text{d}t\\
&=2G \end{aligned}$

$\displaystyle\int_{-\infty}^{+\infty}\dfrac{\text{sech}; x}{x^2+\pi^2}\text{d}x=\dfrac{4}{\pi}-1$

$\begin{aligned} &\int_0^{\infty}\dfrac{\text{d}x}{(x^2+b^2)\cosh ax}\\
=&\dfrac{\pi}{2a}\int_{-\infty}^{+\infty}\dfrac{\text{d}x}{\left(\pi^2\frac{x^2}{a^2}+b^2\right)\cosh \pi x}\\
=&\dfrac{\pi}{2a}\cdot\frac{a}{b}\int_{-\infty}^{+\infty}\dfrac{e^{-2abx}}{\cosh \pi x}\text{d}x\\
=&\dfrac{\pi}{b}\int_0^{\infty}e^{-2abx}\sum_{n=0}^{\infty}2(-1)^ne^{-(2n+1)\pi x}\text{d}x\\
=&\dfrac{2\pi}{b}\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)\pi+2ab}\\
=&\dfrac{1}{2b}\left[\psi\left(\dfrac{ab}{2\pi}+\dfrac34\right)-\psi\left(\dfrac{ab}{2\pi}+\frac14\right)\right] \end{aligned}$

$\displaystyle\int_0^{\infty}\dfrac{\cos tx}{\cosh\dfrac{\pi x}{2}(1+x^2)}\text{d}x$

$\cosh(t)\ln[2\cosh(t)]-t\sinh(t)$

$\displaystyle\int_0^{\infty}\dfrac{1-\cos x}{x(e^x-1)}\text{d}x$

$\begin{aligned} I=&\int_0^{\infty}\dfrac{1-\cos x}{x(e^x-1)}\text{d}x\\
I(a)=&\int_0^{\infty}\dfrac{1-\cos ax}{x(e^x-1)}\text{d}x\\
I'(a)&=\int_0^{\infty}\dfrac{\sin ax}{e^x-1}\text{d}x\\
=&\int_0^{\infty}\sin ax\cdot\sum_{n=1}^{\infty}e^{-nx}\text{d}x\\
=&\sum_{n=1}^{\infty}\int_0^{\infty}\sin ax\cdot e^{-nx}\text{d}x\\
=&\sum_{n=1}^{\infty}\dfrac{a}{n^2+a^2} \end{aligned}$
$\begin{aligned} I=&I(1)=\int_0^{1}\sum_{n=1}^{\infty}\dfrac{a}{n^2+a^2}\\
=&\dfrac12\sum_{n=1}^{\infty}\ln\left(1+\dfrac1{n^2}\right)\\
=&\dfrac12\ln\left(\dfrac{\sinh x}{x}\right) \end{aligned}$

$\displaystyle\dfrac{\sinh x}{x}=\prod_{k=1}^{\infty}\left(1+\dfrac{x^2}{k^2\pi^2}\right)$

$\displaystyle\int_0^{\pi}\dfrac{x}{\left(\sin x+\sqrt{2}\right)^2}\text{d}x=\dfrac{(\pi-2)\pi}{2\sqrt{2}}$

$\begin{aligned} I=&\int_0^{\pi}\dfrac{x}{\left(\sin x+\sqrt{2}\right)^2}\text{d}x\\
=&\dfrac\pi2\int_0^{\pi}\dfrac{\text{d}x}{(\sin x+\sqrt{2})^2}\\
=&\pi\int_0^{\infty}\dfrac{t^2+1}{\left[2t+\sqrt2(t^2+1)\right]^2}\text{d}t\\
=&\dfrac{\pi(\pi-2)}{2\sqrt{2}} \end{aligned}$

$\displaystyle\int_0^{\infty}\dfrac{\text{d}x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}=\dfrac{\pi}{2(1+\sqrt{2})}$

$\begin{aligned} I=&\int_0^{\infty}\dfrac{\text{d}x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\
=&\int_0^{\infty}\dfrac{\frac1{x^2}\text{d}x}{\left[\frac1{x^4}+(1+2\sqrt2)\frac1{x^2}+1\right][\frac1{x^{100}}-\frac1{x^{98}}+\cdots+1]}\\
=&\int_0^{\infty}\dfrac{x^{102}\text{d}x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\
2I=&\int_0^{\infty}\dfrac{1+x^{102}\text{d}x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\
=&\int_0^{\infty}\dfrac{(x^2+1)(x^{100}-x^{98}+\cdots+1)\text{d}x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\
=&\int_0^{\infty}\dfrac{x^2+1}{x^4+(1+2\sqrt{2})x^2+1}\text{d}x\\
=&\int_0^{\infty}\dfrac{1+\frac1{x^2}}{x^2+\frac1{x^2}+(1+2\sqrt{2})}\text{d}x\\
=&\int_0^{\infty}\dfrac{\text{d}(x-\frac1x)}{(x-\frac1x)^2+3+2\sqrt2}\\
=&\dfrac{1}{1+\sqrt2}\arctan\left(\dfrac{t-\frac1t}{1+\sqrt2}\right)\bigg|_0^{\infty}\\
=&\dfrac{\pi}{1+\sqrt2}\\
I=&\dfrac{\pi}{2(1+\sqrt2)} \end{aligned}$

$\begin{aligned} &\displaystyle\int_0^{\frac\pi4}\left[\dfrac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right]x\cdot\exp\left[\dfrac{x^2-1}{x^2+1}\right]\text{d}x\\
&=-\dfrac14\exp\left[-\dfrac{16-\pi^2}{16+\pi^2}\right]\ln\left[\dfrac{16-\pi^2}{16+\pi^2}\right] \end{aligned}$

$\displaystyle\int_0^{+\infty}\dfrac{e^{-x^2}}{(x^2+\frac12)^2}\text{d}x$

$\begin{aligned} &\int_0^{\infty}\dfrac{e^{-x^2}}{(x^2+\frac12)^2}\text{d}x\\
=&4\int_0^{\infty}\dfrac{e^{-x^2}}{(2x^2+1)^2}\text{d}x\\
=&\dfrac{4}{\sqrt{2}}\int_0^{\infty}\dfrac{e^{-\frac{x^2}2}}{(x^2+1)^2}\text{d}x\\
=&\dfrac{4}{\sqrt{2}}\cdot\dfrac{1}{\Gamma(2)}\int_0^{\infty}t\cdot e^{-t}\text{d}t\int_0^{\infty}e^{-\frac{x^2}2-tx^2}\text{d}x\\
=&\dfrac{4\sqrt\pi}{\sqrt{2}}\int_0^{\infty}\dfrac{te^{-t}}{\sqrt{4t+2}}\text{d}t\\
=&\dfrac{4\sqrt\pi}{\sqrt{2}}\cdot\dfrac{1}{2\sqrt{2}}\\
=&\sqrt{\pi} \end{aligned}$