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Math formulas: Exponential definite integrals

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0eaxcosbxdx=aa2+b2 \int^\infty_0 e^{-ax} \cos bx \, dx = \frac{a}{a^2 + b^2}
0eaxsinbxdx=ba2+b2 \int^\infty_0 e^{-ax} \sin bx \, dx = \frac{b}{a^2 + b^2}
0eaxsinbxxdx=arctanba \int^\infty_0 \frac{e^{-ax} \sin bx}{x} \, dx = \arctan \frac{b}{a}
0eaxebxxdx=lnba \int^\infty_0 \frac{e^{-ax}-e^{-bx}}{x} dx = \ln \frac{b}{a}
0eax2dx=12πa \int^\infty_0 e^{-ax^2} \, dx = \frac{1}{2} \sqrt{ \frac{\pi}{a} }
0eax2cosbxdx=12πaeb24a \int^\infty_0 e^{-ax^2} \cos bx \, dx = \frac{1}{2} \sqrt{ \frac{\pi}{a} } e^{-\frac{b^2}{4a}}
e(ax2+bx+c)dx=πaeb24ac4a \int^\infty_{-\infty} e^{-(ax^2+bx+c)} dx = \sqrt{\frac{\pi}{a}} e^\frac{b^2-4ac}{4a}
0xneaxdx=Γ(n+1)an+1 \int^\infty_0 x^n\,e^{-ax}dx = \frac{\Gamma(n+1)}{a^{n+1}}
0xmeax2dx=Γ(m+12)2a(m+1)/2 \int^\infty_0 x^m\,e^{-ax^2}dx = \frac{\Gamma\left(\frac{m+1}{2}\right)}{2a^{(m+1)/2}}
0e(ax2+b/x2)dx=12πae2ab \int^\infty_0 e^{-\left(ax^2+b/x^2\right)} dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}
0xdxex1=π26 \int^\infty_0 \frac{x\,dx}{e^x-1} = \frac{\pi^2}{6}
0xn1ex1dx=Γ(n)(11n+12n+13n+) \int^\infty_0 \frac{x^{n-1}}{e^x-1}dx = \Gamma (n) \left( \frac{1}{1^n} + \frac{1}{2^n} + \frac{1}{3^n} + \cdots \right)
0xdxex+1=π212 \int^\infty_0 \frac{x\,dx}{e^x+1} = \frac{\pi^2}{12}
0xn1ex+1dx=Γ(n)(11n12n+13n) \int^\infty_0 \frac{x^{n-1}}{e^x+1}dx = \Gamma (n) \left( \frac{1}{1^n} - \frac{1}{2^n} + \frac{1}{3^n} - \cdots \right)
0sinmxe2πx1dx=14cothm212m \int^\infty_0 \frac{\sin mx}{e^{2\pi x} - 1} dx = \frac{1}{4} \coth \frac{m}{2} - \frac{1}{2m}
0(11+xex)dxx=γ \int^\infty_0 \left( \frac{1}{1+x} - e^{-x} \right) \frac{dx}{x} = \gamma
0ex2exxdx=12γ \int^\infty_0 \frac{e^{-x^2} - e^{-x}}{x} dx = \frac{1}{2} \gamma
0(1ex1exx)dx=γ \int^\infty_0 \left( \frac{1}{e^x-1} - \frac{e^{-x}}{x} \right) dx = \gamma
0eaxebxxsec(px)dx=12ln(b2+p2a2+p2) \int^\infty_0 \frac{e^{-ax} - e^{-bx}}{x \sec (px)} dx = \frac{1}{2} \ln\left( \frac{b^2+p^2}{a^2+p^2}\right)
0eaxebxxcsc(px)dx=arctanbparctanap \int^\infty_0 \frac{e^{-ax} - e^{-bx}}{x \csc (px)} dx = \arctan \frac{b}{p} - \arctan \frac{a}{p}
0eax(1cosx)x2dx=arccotaa2ln(a2+1) \int^\infty_0 \frac{e^{-ax}(1-\cos x)}{x^2} dx = \mathrm{arccot}\,a - \frac{a}{2} \ln (a^2 + 1)

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