Definite Integrals of Exponential Functions

\( \int^\infty_0 e^{-ax} \cos bx \, dx = \frac{a}{a^2 + b^2} \)

 

\( \int^\infty_0 e^{-ax} \sin bx \, dx = \frac{b}{a^2 + b^2} \)

 

\( \int^\infty_0 \frac{e^{-ax} \sin bx}{x} \, dx = \arctan \frac{b}{a} \)

 

\( \int^\infty_0 \frac{e^{-ax}-e^{-bx}}{x} dx = \ln \frac{b}{a} \)

 

\( \int^\infty_0 e^{-ax^2} \, dx = \frac{1}{2} \sqrt{ \frac{\pi}{a} } \)

 

\( \int^\infty_0 e^{-ax^2} \cos bx \, dx = \frac{1}{2} \sqrt{ \frac{\pi}{a} } e^{-\frac{b^2}{4a}} \)

 

\( \int^\infty_{-\infty} e^{-(ax^2+bx+c)} dx = \sqrt{\frac{\pi}{2}} e^\frac{b^2-4ac}{4a} \)

 

\( \int^\infty_0 x^n\,e^{-ax}dx = \frac{\Gamma(n+1)}{a^{n+1}} \)

 

\( \int^\infty_0 x^m\,e^{-ax^2}dx = \frac{\Gamma\left(\frac{m+1}{2}\right)}{2a^{(m+1)/2}} \)

 

\( \int^\infty_0 e^{-\left(ax^2+b/x^2\right)} dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}} \)

 

\( \int^\infty_0 \frac{x\,dx}{e^x-1} = \frac{\pi^2}{6} \)

 

\( \int^\infty_0 \frac{x^{n-1}}{e^x-1}dx = \Gamma (n) \left( \frac{1}{1^n} + \frac{1}{2^n} + \frac{1}{3^n} + \cdots \right) \)

 

\( \int^\infty_0 \frac{x\,dx}{e^x+1} = \frac{\pi^2}{12} \)

 

\( \int^\infty_0 \frac{x^{n-1}}{e^x+1}dx = \Gamma (n) \left( \frac{1}{1^n} – \frac{1}{2^n} + \frac{1}{3^n} – \cdots \right) \)

 

\( \int^\infty_0 \frac{x\,dx}{e^x+1} = \frac{\pi^2}{12} \)

 

\( \int^\infty_0 \frac{x^{n-1}}{e^x+1}dx = \Gamma (n) \left( \frac{1}{1^n} – \frac{1}{2^n} + \frac{1}{3^n} – \cdots \right) \)

 

\( \int^\infty_0 \frac{\sin mx}{e^{2\pi x} – 1} dx = \frac{1}{4} \coth \frac{m}{2} – \frac{1}{2m} \)

 

\( \int^\infty_0 \left( \frac{1}{1+x} – e^{-x} \right) \frac{dx}{x} = \gamma \)

 

\( \int^\infty_0 \frac{e^{-x^2} – e^{-x}}{x} dx = \frac{1}{2} \gamma \)

 

\( \int^\infty_0 \left(  \frac{1}{e^x-1} – \frac{e^{-x}}{x} \right) dx = \gamma \)

 

\( \int^\infty_0 \frac{e^{-ax} – e^{-bx}}{x \sec (px)}  dx = \frac{1}{2} \ln\left( \frac{b^2+p^2}{a^2+p^2}\right) \)

 

\( \int^\infty_0 \frac{e^{-ax} – e^{-bx}}{x \csc (px)}  dx = \arctan \frac{b}{p} – \arctan \frac{a}{p} \)

 

\( \int^\infty_0  \frac{e^{-ax}(1-\cos x)}{x^2}  dx = \mathrm{arccot}\,a – \frac{a}{2} \ln (a^2 + 1) \)

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