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| \PMlinkid{1.}{7065} \; $\displaystyle\int_0^\infty e^{-x^2}\,dx \;=\; \frac{\sqrt{\pi}}{2}$\\ |
\PMlinkid{1.}{7065} \; $\displaystyle\int_0^\infty e^{-x^2}\,dx \;=\; \frac{\sqrt{\pi}}{2}$\\ |
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| \PMlinkid{2.}{11495} \; $\displaystyle\int_0^\infty e^{-x^2}\cos{kx}\,dx\;=\;\frac{\sqrt{\pi}}{2}e^{-\frac{1}{4}k^2}$\\ |
\PMlinkid{2.}{11495} \; $\displaystyle\int_0^\infty e^{-x^2}\cos{kx}\,dx\;=\;\frac{\sqrt{\pi}}{2}e^{-\frac{1}{4}k^2}$\\ |
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| \PMlinkid{3.}{11504} \; $\displaystyle\int_0^\infty \frac{e^{-x^2}}{a^2\!+\!x^2}\,dx |
\PMlinkid{3.}{11504} \; $\displaystyle\int_0^\infty \frac{e^{-x^2}}{a^2\!+\!x^2}\,dx |
| \;=\;\frac{\pi}{2a}e^{a^2}\,{\rm erfc}\,a$\\ |
\;=\;\frac{\pi}{2a}e^{a^2}\,{\rm erfc}\,a$\\ |
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| \PMlinkid{4.}{10980} \; $\displaystyle\int_0^\infty\sin{x^2}\,dx \;=\; \int_0^\infty\cos{x^2}\,dx \;=\; |
\PMlinkid{4.}{10980} \; $\displaystyle\int_0^\infty\sin{x^2}\,dx \;=\; \int_0^\infty\cos{x^2}\,dx \;=\; |
| \frac{\sqrt{2\pi}}{4}$\\ |
\frac{\sqrt{2\pi}}{4}$\\ |
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| \PMlinkid{5.}{7082} \; $\displaystyle\int_0^\infty\frac{\sin{x}}{x}\,dx \;=\; \frac{\pi}{2}$\\ |
\PMlinkid{5.}{7082} \; $\displaystyle\int_0^\infty\frac{\sin{x}}{x}\,dx \;=\; \frac{\pi}{2}$\\ |
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| \PMlinkid{6.}{11487} \; $\displaystyle\int_0^\infty\left(\frac{\sin{x}}{x}\right)^2 dx \;=\; \frac{\pi}{2}$\\ |
\PMlinkid{6.}{11487} \; $\displaystyle\int_0^\infty\left(\frac{\sin{x}}{x}\right)^2 dx \;=\; \frac{\pi}{2}$\\ |
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| \PMlinkid{7.}{11487} \; $\displaystyle\int_0^\infty\frac{1-\cos{kx}}{x^2}\,dx \;=\; \frac{\pi k}{2}$\\ |
\PMlinkid{7.}{11487} \; $\displaystyle\int_0^\infty\frac{1-\cos{kx}}{x^2}\,dx \;=\; \frac{\pi k}{2}$\\ |
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| \PMlinkid{8.}{11480} \; $\displaystyle\int_0^\infty\frac{x^{-k}}{x\!+\!1}\,dx \;=\; \frac{\pi}{\sin{\pi k}} |
\PMlinkid{8.}{11480} \; $\displaystyle\int_0^\infty\frac{x^{-k}}{x\!+\!1}\,dx \;=\; \frac{\pi}{\sin{\pi k}} |
| \quad (0 < k < 1)$\\ |
\quad (0 < k < 1)$\\ |
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| 9. \; $\displaystyle\int_{-\infty}^\infty\frac{e^{kx}}{1\!+\!e^x}\,dx \;=\; \frac{\pi}{\sin{\pi k}} |
\PMlinkid{9.}{7136} \; $\displaystyle\int_0^\infty\frac{\cos{kx}}{x^2\!+\!1}\,dx \;=\; \frac{\pi}{2e^k}$\\ |
| \quad (0 < k < 1)$\\ |
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| \PMlinkid{10.}{7136} \; $\displaystyle\int_0^\infty\frac{\cos{kx}}{x^2\!+\!1}\,dx \;=\; \frac{\pi}{2e^k}$\\ |
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\PMlinkid{11.}{11489} \; $\displaystyle\int_0^\infty\frac{a\cos{x}}{x^2\!+\!a^2}\,dx
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\PMlinkid{1.}{11489} \; $\displaystyle\int_0^\infty\frac{a\cos{x}}{x^2\!+\!a^2}\,dx
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| \;=\; \int_0^\infty\frac{x\sin{x}}{x^2\!+\!a^2}\,dx \;=\; \frac{\pi}{2e^a} \quad\; (a > 0)$\\ |
\;=\; \int_0^\infty\frac{x\sin{x}}{x^2\!+\!a^2}\,dx \;=\; \frac{\pi}{2e^a} \quad\; (a > 0)$\\ |
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12. \; $\displaystyle\int_0^\infty\frac{\sin{ax}}{x(x^2\!+\!1)}\,dx \;=\; \frac{\pi}{2}(a-e^{-a})? \quad\; (a > 0)$\\
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11. \; $\displaystyle\int_0^\infty\frac{\sin{ax}}{x(x^2\!+\!1)}\,dx \;=\; \frac{\pi}{2}(a-e^{-a})? \quad\; (a > 0)$\\
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\PMlinkid{13.}{9223} \; $\displaystyle\int_0^\infty e^{-x}x^{-\frac{3}{2}}\,dx \;=\; \sqrt{\pi}$\\
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\PMlinkid{12.}{9223} \; $\displaystyle\int_0^\infty e^{-x}x^{-\frac{3}{2}}\,dx \;=\; \sqrt{\pi}$\\
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\PMlinkid{14.}{10637} \; $\displaystyle\int_0^\infty e^{-x}x^3\sin{x}\,dx \;=\; 0$\\
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\PMlinkid{13.}{10637} \; $\displaystyle\int_0^\infty e^{-x}x^3\sin{x}\,dx \;=\; 0$\\
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\PMlinkid{15.}{7891} \; $\displaystyle\int_0^\infty\!\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right) dx \;=\; \gamma$\\
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\PMlinkid{14.}{7891} \; $\displaystyle\int_0^\infty\!\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right) dx \;=\; \gamma$\\
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\PMlinkid{16.}{11516} \; $\displaystyle\int_0^\infty\!\frac{\cos{ax^2}-\cos{ax}}{x} dx \;=\; \frac{\gamma+\ln{a}}{2} \quad (a > 0)$\\
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\PMlinkid{15.}{11516} \; $\displaystyle\int_0^\infty\!\frac{\cos{ax^2}-\cos{ax}}{x} dx \;=\; \frac{\gamma+\ln{a}}{2} \quad (a > 0)$\\
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\PMlinkid{17.}{11511} \; $\displaystyle\int_0^\infty\frac{e^{-ax}\!-\!e^{-bx}}{x}\,dx \;=\; \ln\frac{b}{a} \quad (a > 0,\;\, b > 0)$\\
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\PMlinkid{16.}{11511} \; $\displaystyle\int_0^\infty\frac{e^{-ax}\!-\!e^{-bx}}{x}\,dx \;=\; \ln\frac{b}{a} \quad (a > 0,\;\, b > 0)$\\
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\PMlinkid{18.}{11526} \; $\displaystyle\int_1^\infty\left(\arcsin\frac{1}{x}-\frac{1}{x}\right)\,dx \;=\; 1+\ln{2}-\frac{\pi}{2}$\\
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\PMlinkid{17.}{11526} \; $\displaystyle\int_1^\infty\left(\arcsin\frac{1}{x}-\frac{1}{x}\right)\,dx \;=\; 1+\ln{2}-\frac{\pi}{2}$\\
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