两道有用的定积分（量子力学中的）
本帖最后由 282842712474 于 2015-6-1 09:03 编辑求证
$$\int_0^T \exp\left(-\frac{a}{T-t}-\frac{b}{t}\right)\frac{\dif t}{\sqrt{(T-t)t^3}}=\sqrt{\frac{\pi}{bT}}\exp\left(-\frac{1}{T}(\sqrt{a}+\sqrt{b})^2\right)$$
$$\int_0^T \exp\left(-\frac{a}{T-t}-\frac{b}{t}\right)\frac{\dif t}{\left[\sqrt{(T-t)t}\right]^3}=\sqrt{\frac{\pi}{T^3}}\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}\exp\left(-\frac{1}{T}(\sqrt{a}+\sqrt{b})^2\right)$$
（已经仔细检查过，录入没有错误）
这两个积分在量子力学的微扰论中很有用，可是我想不出积分的思路，求解释~~ \[\begin{array}{l}
I= \int\limits_0^T {\exp \left( { - \frac{a}{{T - t}} - \frac{b}{t}} \right)} \frac{{dt}}{{\sqrt {\left( {T - t} \right){t^3}} }}\left( {replace t with\frac{T}{{x\sqrt {\frac{a}{b}}+ 1}} } \right) \\
= \frac{1}{T}\sqrt{{\frac{a}{b}}}\exp \left( { - \frac{a}{T} - \frac{b}{T}} \right) \times \int\limits_0^\infty{\exp \left( { - \frac{{\sqrt {ab} }}{T}\left( {x + \frac{1}{x}} \right)} \right)\frac{1}{{\sqrt x }}} dx \\
= \frac{1}{T}\sqrt{{\frac{a}{b}}}\exp \left( { - \frac{a}{T} - \frac{b}{T}} \right) \times \int\limits_0^\infty{\exp \left( { - \frac{{\sqrt {ab} }}{T}\left( {x + \frac{1}{x}} \right)} \right)\frac{1}{{\sqrt {{x^3}} }}} dx\left( {replace x with\frac{1}{x} } \right) \\
= \frac{1}{T}\sqrt{{\frac{a}{b}}}\exp \left( { - \frac{a}{T} - \frac{b}{T} - \frac{{2\sqrt {ab} }}{T}} \right) \times \int\limits_0^\infty{\exp \left( { - \frac{{\sqrt {ab} }}{T}{{\left( {\sqrt x- \frac{1}{{\sqrt x }}} \right)}^2}} \right)} d\left( {\sqrt x- \frac{1}{{\sqrt x }}} \right) \\
= \frac{1}{T}\sqrt{{\frac{a}{b}}}\exp \left( { - \frac{a}{T} - \frac{b}{T} - \frac{{2\sqrt {ab} }}{T}} \right) \times \int\limits_{ - \infty }^\infty{\exp \left( { - \frac{{\sqrt {ab} }}{T}{u^2}} \right)} du \\
= \frac{{\sqrt {a\pi } }}{T}\exp \left( { - \frac{a}{T} - \frac{b}{T} - \frac{{2\sqrt {ab} }}{T}} \right) \\
\end{array}\] 试一下复积分，找个积分路径。
页:
[1]