DEFINITE INTEGRALS CONTAINING EXPONENTIAL FUNCTIONS

1.
$\displaystyle\int_{0}^{\infty}e^{-ax}\cos bx dx=\displaystyle \frac{a}{a^2+b^2}$

2.
$\displaystyle\int_{0}^{\infty}e^{-ax}\sin bx dx=\displaystyle \frac{b}{a^2+b^2}$

3.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-ax}\sin bx}{x}dx=\tan^{-1}\displaystyle \frac{b}{a}$

4.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-ax}-e^{-bx}}{x}dx=\ln\displaystyle \frac{b}{a}$

5.
$\displaystyle\int_{0}^{\infty}e^{-ax^2}dx=\displaystyle \frac{1}{2}\displaystyle \sqrt{\displaystyle \frac{\pi}{a}}$

6.
$\displaystyle\int_{0}^{\infty}e^{-ax^2}\cos bx dx=\displaystyle \frac{1}{2}\displaystyle \sqrt{\displaystyle \frac{\pi}{a}}e^{-b^2/4a}$

7.
$\displaystyle\int_{0}^{\infty}e^{-(ax^2+bx+c)}dx=\displaystyle \frac{1}{2}\displaystyle \sqrt{\displaystyle \frac{\pi}{a}}e^{(b^2-4ac)/4a}$

8.
$\displaystyle\int_{-\infty}^{\infty}e^{-(ax^2+bx+c)}dx=\displaystyle \sqrt{\displaystyle \frac{\pi}{a}}e^{(b^2-4ac)/4a}$

9.
$\displaystyle\int_{0}^{\infty}x^n e^{-ax}dx=\displaystyle \frac{\Gamma(n+1)}{a^n+1}$

10.
$\displaystyle\int_{0}^{\infty}x^m e^{-ax^2}dx=\displaystyle \frac{\Gamma[(m+1)/2]}{2a^{(m+1)/2}}$

11.
$\displaystyle\int_{0}^{\infty}e^{-(ax^2+b/x^2)}dx=\displaystyle \frac{1}{2}\displaystyle \sqrt{\displaystyle \frac{\pi}{a}}e^{-2\displaystyle \sqrt{ab}}$

12.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{xdx}{e^x-1}=\displaystyle \fr...
...3^2}+\displaystyle \frac{1}{4^2}+\cdot\cdot\cdot =\displaystyle \frac{\pi^2}{6}$

13.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{x^{n-1}}{e^x-1}dx=\Gamma(n+1)...
...\displaystyle \frac{1}{2^n}+\displaystyle \frac{1}{3^n}+\cdot\cdot\cdot \right)$

14.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{xdx}{e^x+1}=\displaystyle \fr...
...2}-\displaystyle \frac{1}{4^2}+
\cdot\cdot\cdot =\displaystyle \frac{\pi^2}{12}$

15.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{x^{n-1}}{e^x+1}dx=\Gamma(n+1)...
...\displaystyle \frac{1}{2^n}+\displaystyle \frac{1}{3^n}-\cdot\cdot\cdot \right)$

16.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{\sin mx}{e^{2\pi x}-1}dx=\displaystyle \frac{1}{4}\coth\displaystyle \frac{m}{2}-\displaystyle \frac{1}{2m}$

17.
$\displaystyle\int_{0}^{\infty}\left( \displaystyle \frac{1}{1+x}-e^{-x}\right)\displaystyle \frac{dx}{x}=\gamma$

where the constant $\gamma$ is the Euler's constant.

18.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-x^2}-e^{-x}}{x}dx=\displaystyle \frac{1}{2}\gamma$

where the constant $\gamma$ is the Euler's constant.

19.
$\displaystyle\int_{0}^{\infty}\left( \displaystyle \frac{1}{e^x-1}-\displaystyle \frac{e^{-x}}{x}\right)dx=\gamma$

where the constant $\gamma$ is the Euler's constant.

20.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-ax}-e^{-bx}}{x\sec px}dx=\displaystyle \frac{1}{2}\ln\left(\displaystyle \frac{b^2+p^2}{a^2+p^2}\right)$

21.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-ax}-e^{-bx}}{x\csc px}dx=\tan^{-1}\displaystyle \frac{b}{p}-\tan^{-1}\displaystyle \frac{a}{p}$

22.
$\displaystyle\int_{0}^{\infty}\displaystyle \frac{e^{-ax}(1-\cos x)}{x^2}dx=\cot^{-1}a-\displaystyle \frac{a}{2}\ln(a^2+1)$

[Tables]

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