The Integral Corner

This section is dedicated to the pleasure of calculating integrals. Have a look at some integrals I find most interesting.

Arctangent’s in the Spotlight

  1. \displaystyle \int_1^{\infty} \frac{\pi-2\arctan{x}}{x}\text{dx}=2G where G is Catalan’s constant.
  2. \displaystyle \int_0^{\infty}\frac{\arctan(\arctan(\ln{x}))}{(\ln^{2}{x}+1)(x^2+1)}dx=0=\int_0^{\infty}\frac{x^2\text{tanh}(\sin(\arctan(\ln^{3}{x})))}{(x^4-x^2+1)(\ln^{6}{x}+\ln^{4}{x}+\ln^{2}{x}+1)(x^2+x+1)}dx
  3. For any q>1 and m>0, \displaystyle \int_{-m}^{m}\frac{\text{sech}(x\ln{q})}{2\arctan(q^{x})\arctan(q^{-x})+\pi+2}dx=\frac{4}{(\pi+4)\ln{q}}\ln\left(\frac{\arctan(q^{m})+1}{\arctan(q^{-m})+1}\right).
  4. For any n \in \mathbb{N}, \displaystyle \int_0^{1} \int_0^{x_{n-1}} \cdots \int_0^{x_2} \int_0^{x_1} \frac{\arctan{x_0}}{x_0 x_1 x_2 \cdots x_{n-2} x_{n-1}}dx_0 dx_1 dx_2 \cdots dx_{n-2} dx_{n-1} =\sum_{r=1}^{\infty} \frac{(-1)^{r-1}}{(2r-1)^{n+1}}.
  5. For any a,b \in \mathbb{R} for which a/b>0, \displaystyle    \int_0^{\infty}\frac{1}{x}\left(\arctan{\frac{1}{ax}}-\arctan{\frac{1}{bx}}\right)dx=\frac{\pi}{2}\ln{\frac{b}{a}}.
  6. For any a,b\in\mathbb{R} with a/b>0, \displaystyle  \int_0^{\infty}\frac{1}{x} \left(\arctan \left(\frac{\sin(ax)}{ax}\right)-\arctan\left(\frac{\sin(bx)}{bx}\right)\right)dx=\frac{\pi}{4}\ln\frac{b}{a}.
  7. For any b,q,p for which b>0 and \text{sgn}(p)=\text{sgn}(q)\neq 0, \displaystyle \int_0^{\infty} \frac{\arctan\left(\frac{px^b}{q}\right)\arctan\left(\frac{q}{px^b}\right)}{q^2x^{1-b}+p^2x^{b+1}}dx=\frac{\pi^3}{48pbq}.
  8. For any a,b \in \mathbb{R} for which a/b>0 and n \in \mathbb{N}, \displaystyle \int_0^{\infty}\frac{1}{x}\left(\arctan^n(ax)-\arctan^n(bx)\right)dx=\left(\frac{\pi}{2}\right)^n\ln{\frac{a}{b}}.
  9. For any n,m\in\mathbb{N} with n\geq m\geq 2, \displaystyle \int_0^{\infty}\frac{\arctan(x^n)}{x^m}dx=\frac{\pi}{2(m-1)\cos\left(\frac{m-1}{2n}\pi\right)}.
  10. For any a\in\mathbb{R}\backslash{0}, \displaystyle \int_0^{\infty}\frac{\arctan{(x)}(1+x^2)}{(x^2+a^2)(x^2+a^{-2})}dx=\frac{a\pi^2}{4(a^2+1)}=\int_0^{\infty}\frac{\arctan{(1/x)}(1+x^2)}{(x^2+a^2)(x^2+a^{-2})}dx.

For the love of Euler’s Constant

  1. \displaystyle \int_{e^{4/\pi}}^{\infty}\frac{\log{x}}{x^2+e^{8/\pi}}dx=\frac{G+1}{e^{4/\pi}}
  2. \displaystyle \int_0^{\infty}\frac{dx}{x^3(e^{\pi/x}+1)}=\frac{1}{12}
  3. \displaystyle \int_1^2 \left(2\lfloor x\rfloor +\frac{1}{\lfloor x \rfloor +\frac{1}{2\lfloor x\rfloor +\frac{2}{3\lfloor x\rfloor+\frac{3}{4\lfloor x \rfloor +\frac{4}{5\lfloor x\rfloor+\cdots} }}}}\right)dx=e
  4. \displaystyle \int_0^{\infty} \{x\}e^{-\lfloor x\rfloor}dx =\frac{e}{2(e-1)}
  5. Suppose T is an n \times n complex matrix whose real part is positive definite. If \langle \cdot, \cdot \rangle denotes the standard complex inner product then \displaystyle \int_{-\infty}^{\infty} \exp\left(-\langle y, Ty \rangle \right) dy=\frac{\pi^{n/2}}{(\det{T})^{1/2}} where the principal value of the square root is taken. (See A Brief Introduction to Theta Functions by Richard E. Bellman)
  6. For any b>0>a, \displaystyle \int_{-\infty}^{\infty} \frac{dx}{e^{ax}+e^{bx}}=\frac{\pi}{(b-a)\sin\left(\frac{b\pi}{b-a}\right)}.
  7. For any a,b>0, \displaystyle \int_0^{\infty}e^{-bx^a}dx=\frac{1}{ab^{1/a}}\Gamma\left(\frac{1}{a}\right).
  8. For any a,b>0, \displaystyle \int_{\mathbb{R}^2}e^{-b(x^2+y^2)}\cosh(ax)\cos(ay)dxdy=\frac{\pi}{b}
  9. For any \alpha,\beta,\gamma,\delta>0, \displaystyle \int_0^{\infty}x^{\alpha}\delta^{-\beta x^{\gamma}}dx=\frac{1}{\gamma}\left(\beta\log{\delta}\right)^{-\frac{\alpha+1}{\gamma}}\Gamma\left(\frac{\alpha+1}{\gamma}\right)
  10. For any a,b,\delta>0 and n,m\in\mathbb{N}, \displaystyle \int_0^{\infty}\frac{\delta^{-ax^n}-\delta^{-bx^n}}{x^m}dx=\frac{1}{m-1}\Gamma\left(\frac{n-m+1}{n}\right)\left(b^{\frac{m-1}{n}}-a^{\frac{m-1}{n}}\right)(\log{\delta})^{\frac{m-1}{n}}

Logs. Everywhere.

  1. \displaystyle \int_0^{\pi}\log\left(\frac{1-\cos{x}}{x^2}\right)dx=-\pi\left(\log\left(2\pi^2\right)-2\right)
  2. \displaystyle \int_0^{\pi/4}\log\left(1-\tan^2{x}\right)\mathrm{d}x=\frac{\pi}{4}\log{2}-G
  3. \displaystyle \int_0^{\pi /4} \log(\cot{x}-\tan{x})dx=\frac{\pi}{4}\log{2}
  4. \displaystyle \int_0^1 \frac{\log{x}}{x^2+1}dx=-G
  5. \displaystyle \int_0^1 \log(x)\log(1-x)dx=2-\frac{\pi^2}{6}
  6. \displaystyle \int_0^{\infty} \log\left(1+\frac{2}{x^2}+\frac{1}{4x^4}\right)\log\left(\frac{2x^2+2-\sqrt{3}}{2x^2+2+\sqrt{3}}\right)dx=\frac{4\pi}{\sqrt{2}}\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\log{2}
  7. \displaystyle \int_0^{\pi/4}\log\left(\frac{\sqrt{1+\sec(2x)}}{\sin^3{x}+\sin^2{x}\cos{x}+\sin{x}\cos^2{x}+\cos^3{x}}\right)dx=\frac{\pi}{8}\log{2}
  8. \displaystyle \int_0^{1}\left\lfloor \ln\left(\frac{1}{x}\right) \right\rfloor dx=\frac{1}{e-1}
  9. For any a>0, \displaystyle \int_0^{\pi/2} \ln(a\tan(x))\ln(a\sin(2x))dx=\frac{\pi}{2}\ln(a)\ln\left(\frac{a}{2}\right)
  10. For a>0, \displaystyle \int_0^a \left\lfloor\ln\left(\frac{a}{x}\right)\right\rfloor\left(\left(\log\left(\frac{a}{x}\right)\right)^{-1}+\left(\log\left(\frac{a}{x}\right)\right)^{-2}\right) dx= a\log\left(\frac{e}{e-1}\right)

Radical Integrals

  1. Let f be a real-valued differentiable function on \mathbb{R} that is non-negative and monotonically increasing. Then \displaystyle \int\frac{f'(x)^{\frac{2}{3}}}{1+f(x)^{5/3}}\sqrt{f(x)\sqrt{f'(x)\sqrt{f(x)\sqrt{f'(x)\sqrt{\cdots}}}}}\text{ }\mathrm{d}x=\frac{3}{5}\log\left(1+f(x)^{\frac{5}{3}}\right)+C
  2. Let I\subset\mathbb{R} be open. Suppose f:I\to (0,1) is a continuously differentiable function that is monotone increasing and onto. Then \displaystyle  \int_I\arctan(f(x))\sqrt{f(x)f'(x)\sqrt{\frac{f'(x)}{f(x)^2}\sqrt{\frac{f'(x)}{f(x)^3}\sqrt{\frac{f'(x)}{f(x)^4}\sqrt{\frac{f'(x)}{f(x)^5}\sqrt{\cdots}}}}}} \mathrm{d}x=G where G denotes Catalan’s constant
  3. Let a,b,c,d>0 with ab>1. Then, \displaystyle \int_{c}^{d}\frac{(e^{4x}+3)e^{3x}b^{\frac{e^{3x}-1}{e^{4x}-1}}}{(e^{4x}-1)^2}\sqrt[e^{x}]{a\sqrt[e^{3x}]{b\sqrt[e^x]{a\sqrt[e^{3x}]{b\sqrt[e^{x}]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{(ab)^{\frac{e^{c}}{e^{2c}-e^{-2c}}}-(ab)^{\frac{e^{d}}{e^{2d}-e^{-2d}}}}{\log(ab)}.
  4. Consider the following functions for x>1 : \displaystyle \phi(x)=\sinh\frac{1}{x}, \displaystyle \psi(x)=\cosh\frac{1}{x}-1, and \displaystyle \Lambda(x)=\phi(x)^2+\psi(x)^2. If Q denotes the unique solution to the equation \Lambda(x)=1 over (1,\infty), then \displaystyle \int_Q^{\infty}\log\left(\Lambda(x)\right)\log\sqrt[\frac{x}{2}]{e^{-\frac{1+\psi(x)}{x^2}}\sqrt[2x]{e^{\frac{-\phi(x)}{x^2}}\sqrt[3x]{e^{-\frac{1+\psi(x)}{x^2}}\sqrt[4x]{e^{-\frac{\phi(x)}{x^2}}\sqrt[5x]{\cdots}}}}}\mathrm{d}x=1
  5. Let P and Q real satisfy 0< P\leq Q. Consider the function \displaystyle a(x):=1+e\gamma(x+1,1) for x\geq 0 where \gamma(u,v) is the incomplete gamma function with integral representation \displaystyle \gamma(u,v)=\int_0^{v}t^{u-1}e^{-t}\mathrm{d}t. Then, \displaystyle \int_P^Q\Lambda(x)\log\sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{1}{4e}\log\left(H(P,Q)\right) with \displaystyle \Lambda(x):=\int_0^{1}t^{x}e^{-t}\log(t)\text{ }\mathrm{d}t and \displaystyle H(P,Q):=\frac{a(Q)^{2a(Q)^2}}{a(P)^{2a(P)^2}}\exp(a(P)^2-a(Q)^2).

Jus’ be Cos…

  1. \displaystyle \int_0^{2\pi}\cos(\cos{x})dx=2\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{4^n(n!)^2}=2\pi J_0(1)
  2. \displaystyle \int_0^{1}\cos(x-x^2)dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+1)(4n)(4n-1)\cdots (2n+2)(2n+1)}
  3. \displaystyle \int_0^{\infty}\frac{(1-\cos(2x))\cos(\sin{x})}{x^2}dx=\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}=\pi J_0(2)
  4. For any n \in \mathbb{N}, \displaystyle \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{(\cos(nx)-\cos(nz))(\cos(ny)-\cos(nz))}{(\cos{x}-\cos{z})(\cos{y}-\cos{z})}dxdydz=\pi^3 n
  5. For any a>0, \displaystyle \int_0^{\infty}\left(1-\cos\left(\frac{1}{a+x^2}\right)\right)dx=\frac{2\pi}{\sqrt{a}} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k)!}\binom{4k-2}{2k-1}\left(\frac{1}{16a^2}\right)^k
  6. For any a>0 and 0<b<1, \displaystyle \int_{0}^{\pi} \frac{\cos(ax)}{b^2+1-2b\cos{x}}dx=\frac{1}{1-b^2}\left(\pi b^a-\sin(a\pi)L(a,b)\right) where \displaystyle L(a,b)=\sum_{k=0}^{\infty} \sum_{m=0}^k \binom{k}{m} \frac{(-1)^m(1-b)^{k-m}}{m+a+1}-\sum_{k=0}^{\infty} \frac{(-1)^kb^{k+1}}{k+a+1}.
  7. For any a,b,s \in \mathbb{R} for which a/b>0, \displaystyle \int_0^{\infty}\frac{1}{x}\left(\cos(ax+s)-\cos(bx+s)\right)dx=\cos(s)\ln{\frac{b}{a}}
  8. For any b,p>0 with p \notin \mathbb{Z} and c \in \mathbb{R}, \displaystyle \int_0^{\infty} \frac{\cos(bx+c)}{x^p}dx=\frac{b^{p-1}\pi}{\Gamma(p)\sin(p\pi)}\sin\left(\frac{p\pi}{2}-c\right)
  9. Let \{d_i\} be a real sequence consisting of n, nonzero, distinct terms. Let s \in \mathbb{R} and a>0 be given. Then \displaystyle \int_{-\infty}^{\infty}\frac{\cos(ax+s)}{\prod_{i=1}^{n} (x^2+d_i^2)}dx=\pi a^{-2}\cos(s)\sum_{k=1}^{n}\frac{c_k e^{-ad_k}}{d_k} where \displaystyle c_k=\prod_{i=1 , i \neq k}^{n}\frac{1}{d_i^2-d_k^2}
  10. For any a,b,c,d \in \mathbb{R} with |a|>|b|+|c|>0, \displaystyle \int_0^{2\pi} \frac{\cos(d\cos{x})\cosh(d\sin{x})}{a+b\cos{x}+c\sin{x}}dx=\frac{2\pi}{\sqrt{a^2-b^2-c^2}}\cosh(c\psi)\cos(b\psi) where \displaystyle \psi=\frac{d\left(a-\sqrt{a^2-b^2-c^2}\right)}{c^2+b^2}.

Spot the \pi

  1. \displaystyle \int_{-\infty}^{\infty} \frac{\sin{x}}{xe^x\cosh{x}}dx=\pi
  2. \displaystyle \int_0^{\infty} \log\left(1+\frac{2}{x^2}+\frac{1}{4x^4}\right)\log\left(\frac{2x^2+2-\sqrt{3}}{2x^2+2+\sqrt{3}}\right)dx=\frac{4\pi}{\sqrt{2}}\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\log{2}
  3. \displaystyle \int_{0}^1 \frac{\log\left(1+x^{\pi^2/12}\right)}{x}dx=1
  4. \displaystyle \sum_{k=0}^{\infty}\left((-1)^{k}\int_{1/e^{1/\sqrt{3}}}^{e^{1/\sqrt{3}}}\frac{3\ln^{2k}{x}}{\pi x}dx\right)=1
  5. \displaystyle \int_{-\infty}^{\infty} \left(\frac{\cosh{x^2}-\sinh{x^2}}{\cosh{x^2}+\sinh{x^2}}\right)^{\pi/2} dx=1
  6. \displaystyle \frac{\partial}{\partial y}\biggr\rvert_{y=1}\left(\int_0^{\infty}\frac{\cosh{ x}\arctan(y\sinh{x})}{\pi^2(1+y^2\sinh^2{x})}dx\right)=-\frac{1}{4}
  7. \displaystyle \int_1^{\infty}\frac{\log\left(\log(x)\right)^{\pi^{-1}}}{x\left(\log^2{x}+1\right)\left(\log^2{x}+2\right)^2}dx=-\frac{1}{16\sqrt{2}}\log\left(\frac{32}{e^2}\right)
  8. For any a>0, \displaystyle \int_a^{\infty} \frac{\ln{x}}{x^2+a^2}dx=\frac{1}{a}\ln\left(a^{\pi/4}e^G\right)
  9. For any a>0, \displaystyle \int_0^{\infty} e^{-ax^2}\ln{x} dx=-\sqrt{\frac{\pi}{a}}\left(\frac{\ln{a}}{4}+\frac{\gamma+2\ln{2}}{4}\right)
  10. Let n \in \mathbb{N} and define f:\mathbb{R} \to \mathbb{R} by f(u)=\sum_{m=0}^{n} u^m. If a,b \in \mathbb{R} satisfy a/b >0, then \displaystyle \int_0^{\infty}\frac{1}{x}\left(\arcsin\left(\frac{f(\sin(ax))}{f(ax)}\right)-\arcsin\left(\frac{f(\sin(bx))}{f(bx)}\right)\right)dx=\frac{\pi}{2}\ln{\frac{b}{a}}

Some Building Blocks

  1. For any a>0 and natural k\geq 1,\displaystyle \int_0^{\infty}\frac{\log{x}}{(a+x^2)^k}dx=\pi (2\sqrt{a})^{1-2k}\left( \binom{2k-2}{k-1}\log(\sqrt{a})-\sum_{j=1}^{k-1}\binom{2k-2-j}{k-1}\frac{2^j}{j}\right)
  2. For any k,n \in \mathbb{Z}_{\geq 0} and 0<r<1, \displaystyle \int_0^{\pi} \frac{\cos^{2k+1}(nx)}{1+r^2-2r\cos{x}}dx=\frac{2^{-2k}\pi r^{n(2k+1)}}{1-r^2}\sum_{m=0}^{k} \begin{pmatrix} 2k+1 \\ m \end{pmatrix} r^{-2nm}
  3. For any k,n \in \mathbb{Z}_{\geq 0} and 0<r<1, \displaystyle \int_0^{\pi} \frac{\cos^{2k}(nx)}{1+r^2-2r\cos{x}}dx=\frac{2^{-2k}\pi}{1-r^2}\left(2\sum_{m=0}^{k-1} \begin{pmatrix} 2k \\ m \end{pmatrix} r^{-2n(m-k)}+\begin{pmatrix}2k \\ k \end{pmatrix}\right)
  4. For any a>0,b\geq 1 and n\in\mathbb{Z}_{\geq 0}, \displaystyle \int_0^1\log(x+b)\log^n(ax)\mathrm{d}x=\sum_{k=0}^n\frac{n!(-1)^k\log^{n-k}(a)}{(n-k)!}\left(\log(b)-(1+b)\text{Li}_1\left(-\frac{1}{b}\right)-(k+1)-b\sum_{r=2}^{k+1}\text{Li}_{r}\left(-\frac{1}{b}\right)\right)
  5. For any a>0,b\geq 1 and n\in\mathbb{Z}_{\geq 0}, \displaystyle \int_0^1\log\left(\frac{b+x}{b-x}\right)\log^n(ax)\mathrm{d}x=\sum_{k=0}^n\frac{n!(-1)^k\log^{n-k}(a)}{(n-k)!}\left(\log\left(\frac{b+1}{b-1}\right)-b\sum_{r=1}^{k+1}2^{1-r}\text{Li}_r\left(\frac{1}{b^2}\right)\right)
  6. For any a,b,c\in\mathbb{R} such that a>-1, b>0 and c>\frac{a+1}{2}, \displaystyle \int_{0}^{\infty}x^a\left(1-\cos\left(\frac{1}{x^c+b}\right)\right)\mathrm{d}x=\frac{2k^{\frac{a+1}{c}}}{c}\Gamma\left(\frac{a+1}{c}\right)\sum_{j=1}^{\infty}\frac{(-1)^{j-1}j\Gamma\left(2j-\frac{a+1}{c}\right)}{{(2j)!}^2k^{2j}}
  7. For any n,m\in \mathbb{N} with m\geq 2 \displaystyle \int_0^{1/2}\frac{1}{(1-x)^{2m}}\ln^n\left(\frac{x}{1-x}\right)\mathrm{d}x=(-1)^nn!\sum_{r=0}^{2(m-1)} \begin{pmatrix} 2(m-1) \\ r \end{pmatrix} \frac{1}{(r+1)^{n+1}}
  8. For any a>0 and n \in \mathbb{N}, \displaystyle \int_0^{\infty}x^{-n-1}\ln\left(1+\exp\left(-\frac{a}{x}\right)\right)\mathrm{d}x=a^{-n}(n-1)!(1-2^{-n})\zeta(n+1)
  9. For any n\in\mathbb{N} such that n\geq 2, \displaystyle \int_0^1x^{n-1}\left\{\frac{1}{x}\right\}\mathrm{d}x=\frac{1}{n-1}-\frac{1}{n}\zeta(n) where \{w\} denotes the fractional part of w\in\mathbb{R}.
  10. Let a,b satisfy -\infty<a\leq b<\infty and n\in \mathbb{N}. Define a finite set of polynomials \displaystyle \left\{P_{i}(x)=\sum_{m=0}^{k_i} e_{m}^{(i)}x^m:x, e_m^{(i)}\in \mathbb{R}; k_i \in \mathbb{Z}_{\geq 0};i=0,1,\cdots,n\right\}. Suppose we have a finite set of real-valued continuous functions on \mathbb{R}, \displaystyle \left\{f_i(x):i=0,1,\cdots,n\right\} and let, for any s_i\in\mathbb{Z}_{\geq 0}, \displaystyle I(a,b,s_0,s_1,\cdots,s_n)=\int_a^b\prod_{i=0}^{n}(f_i(x))^{s_i}\mathrm{d}x. Then,

\displaystyle \int_a^b \prod_{i=0}^n (P_{i}\circ f_i)(x)\mathrm{d}x=\sum_{w_n=0}^{k_n}\cdots\sum_{w_0=0}^{k_0}\left(\prod_{j=0}^{n}e_{w_j}^{(j)}\right)I(a,b,w_0,\cdots,w_n).

Who Said Infinity Can’t Be Fun?

  1. For any constants b_1,b_2,\cdots,b_n>0, a_1,a_2,\cdots,a_n\in\mathbb{R} and function g:\mathbb{R}\to\mathbb{R} for which g\in L_{loc}^{1}(\mathbb{R}), xg\in L^1(\mathbb{R}) and g(x)=-g(-x) for almost every x\in \mathbb{R}, there holds \displaystyle \int_{-\infty}^{\infty}\log\left(\prod_{k=1}^n(1+b_k^x)^{a_k}\right)g(x)\mathrm{d}x=\log\left(\prod_{k=1}^nb_k^{a_k}\right)\int_0^{\infty}xg(x)\mathrm{d}x
  2. Let \{d_i\} be a real sequence consisting of n, nonzero, distinct terms. Let functions f,g:\mathbb{R} \to \mathbb{R} satisfy, for each k=1,\cdots,n, \displaystyle \int_0^{\infty}\frac{f(x)}{x^2+d_k^2}dx=g(d_k). Then \displaystyle \int_{0}^{\infty}\frac{f(x)}{\prod_{i=1}^{n} (x^2+d_i^2)}dx=\sum_{k=1}^{n}c_kg(d_k) where \displaystyle c_k=\prod_{i=1 , i \neq k}^{n}\frac{1}{d_i^2-d_k^2} for k=1,\cdots,n.
  3. Let s,b,k,q \in \mathbb{R} with b>0 and k/q>0. Then, \displaystyle \int_0^{\infty}\int_{-\infty}^{\infty}\frac{\cos((q+k)mu+s)\sin((q-k)mu)}{m\exp({bu^2})}dudm=\frac{\sin(s)}{2}\sqrt{\frac{\pi}{b}}\ln\frac{k}{q}
  4. For any b,p,n>0, with p \notin \mathbb{Z} and h \in \mathbb{R} \displaystyle \int_0^{\infty} \frac{\sin(bx^{n+1}+h)}{x^{n(p-1)+p}}dx=\frac{b^{p-1}\pi}{(n+1)\Gamma(p)\sin(p\pi)}\cos\left(\frac{p\pi}{2}-h\right)
  5. For any a,b>0 and natural k such that b-a=1 and k\geq 2, \displaystyle \int_1^{\infty}\frac{\log\left(\log{x}\right)}{x\left(\log^2{x}+a\right)\left(\log^2{x}+b\right)^k}dx=\frac{\pi}{2}\log\left(\frac{{\sqrt{a}}^{\left(\sqrt{a}\right)^{-1}}}{{\sqrt{b}}^{\left(\sqrt{b}\right)^{-1}}}\right)+\pi\sum_{r=2}^{k} (2\sqrt{b})^{1-2r}\left(\sum_{j=1}^{r-1}\binom{2r-2-j}{r-1}\frac{2^j}{j}-\binom{2r-2}{r-1}\log\sqrt{b}\right)
  6. For any p,q,k for which p,q \geq -\frac{1}{2} and 1<k<2, \displaystyle \int_0^{\infty} \frac{\cos[(p+q+1)x]\sin[(p-q)x]}{x^k}dx=\frac{\pi\left[(2p+1)^{k-1}-(2q+1)^{k-1}\right]}{4(k-1)\Gamma (k-1)\cos{\frac{(k-1)\pi}{2}}}
  7. For any m,s\in \mathbb{R}, \displaystyle \int_{-\infty}^{\infty}\log\left(1+e^{-x^2}\right)\cos(mx+s)\mathrm{d}x=\sqrt{\pi}\cos(s)\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k\sqrt{k}}e^{-\frac{m^2}{4k}}
  8. For any a \in \mathbb{R}\backslash\{0\} and n \in \mathbb{N}, \displaystyle \int_0^{\infty} \frac{x^{4n-2}}{(x^{2n}+a^{2n})^2}dx=\frac{\pi (2n-1)}{4n^2 |a| \sin\frac{\pi}{2n}}
  9. For any a>0, \displaystyle \int_a^{\infty}\frac{\log(x^2+a^2)}{x^2+a^2}\mathrm{d}x=\frac{1}{a}\log\left((2a)^{\pi/2}e^{G}\right)
  10. Let J_0 denote the zeroth order Bessel function of the firtst kind. Then, for any t\in\mathbb{R}, \displaystyle \int_0^{\infty}\left(1-J_0\left(e^{-x^2}\right)\right)x\sin(tx)dx=\frac{t\sqrt{\pi}}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(2k)^{-3/2}}{(k!2^k)^2}\exp\left(-\frac{t^2}{8k}\right)

Complex Analysis to the Rescue!

  1. For |a|>|b|+|c|>0 where a,b,c \in \mathbb{R}, and n \in \mathbb{Z}_{\geq 0} \displaystyle \int_0^{2\pi}\frac{\cos(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \cos(n\phi) and \displaystyle \int_0^{2\pi}\frac{\sin(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \sin(n\phi) where \displaystyle \lambda (a,b,c,n)=\frac{(\text{sgn}(a))^n\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^n}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^n}} and \displaystyle \phi=\text{arg}(b+ic)\in[-\pi,\pi).
  2. For any a,b,c,d,e,f\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0} such that c>|a|+|b|>0 and m\geq 1, \displaystyle \int_{0}^{2\pi}\frac{d\cos(nx)+e\sin(nx)+f}{a\cos(mx)+b\sin(mx)+c}\mathrm{d}x=\frac{2\pi }{m\sqrt{c^2-a^2-b^2}}\sum_{k=0}^{m-1}\left(f+\sqrt{d^2+e^2}R\cos\left(\beta-n\alpha_k\right)\right) where R=\left(\frac{c-\sqrt{c^2-a^2-b^2}}{\sqrt{a^2+b^2}}\right)^{\frac{n}{m}},\text{ } \beta=\text{arg}(d+ie)\in(-\pi,\pi], and, for k\in\{0,\cdots,m-1\}, we define \alpha_k=\frac{\theta+(2k+1)\pi}{m} \text{ with } \theta=\text{arg}(a+bi) \in (-\pi,\pi].
  3. Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m\in\mathbb{Z}_{\geq 0}, for all z\in\mathbb{R} there holds \displaystyle \int_0^{2\pi}\int_0^{2\pi}\frac{(\cos(mx)-\cos(my))(\cos(my)-\cos(mz))}{(a+b\cos(x)+c\sin(x))(a+b\cos(y)+c\sin(y))}\mathrm{d}x\mathrm{d}y=\frac{2\pi^2}{a^2-b^2-c^2}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right) Consequently, there holds for all n\in\mathbb{N} and x_{2n+1}\in\mathbb{R} \displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n-1}\mathrm{d}x_{2n}= \frac{2^n\pi^{2n}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n.
  4. Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0}, for all x_{2n+2}\in\mathbb{R} there holds \displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n+1}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n}\mathrm{d}x_{2n+1}= \Gamma(a,b,c,n,m)\left(\lambda(a,b,c,m)\cos(m\phi)-\lambda(a,b,c,0)\cos\left(mx_{2n+2}\right)\right) where \displaystyle \Gamma(a,b,c,n,m):=\frac{2^{n+1}\pi^{2n+1}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n, \displaystyle \lambda (a,b,c,w):=\frac{\text{sgn}(a)\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^w}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^w}}, and \phi:=\text{arg}(b+ic)\in[-\pi,\pi).
  5. For any b>0 and n,m \in \mathbb{Z}_{\geq 0} with n \geq m+2, \displaystyle \int_0^{\infty}\frac{x^m}{x^n+b}dx=\frac{\pi b^{\frac{m+1-n}{n}}}{n\sin\frac{(m+1)\pi}{n}}.
  6. For any a,b\in\mathbb{R}, \displaystyle \int_{-\pi}^{\pi}e^{be^{a\cos{x}}\cos(a\sin{x})}\cos(be^{a\cos{x}}\sin(a\sin{x})-x)\mathrm{d}x=2\pi abe^b
  7. For any a>0 and m\in\mathbb{N} with m\geq 2, \displaystyle \int_0^{\infty} \frac{\log{x}}{a+x^m}dx=\frac{\pi \csc\left(\frac{\pi}{m}\right)}{a^{\frac{m-1}{m}}m^2}\left(\log{a}-\pi \cot\left(\frac{\pi}{m}\right)\right)

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