我在（0，無(wú)限）限制內(nèi)積分時(shí)遇到困難。我正在嘗試做這個(gè) Jupyter 筆記本（不知道這是否是相關(guān)信息）。代碼如下：import numpy as npimport matplotlib.pyplot as pltimport scipy as scimport mathfrom scipy.integrate import quaddef integrand3(x,z,m,n):    return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))def IgK0(z,m,n):    IntK1 = quad(integrand3, 0, 1, args=(z,m,n))    return IntK1IgK0(1,1,1)這給出了結(jié)果：(0.19224739693489237, 2.1343748650136926e-15)這沒關(guān)系。但是當(dāng)我用無(wú)窮大替換上限時(shí)，我得到“nan”作為輸出。請(qǐng)看下面的代碼：def IgKI(z,m,n):    IntKI = quad(integrand3, 0, np.inf, args=(z,m,n))    return IntKI當(dāng)我使用 (0,0) 值作為 (m,n) 時(shí)，盡管存在一些我不明白的錯(cuò)誤，但至少我得到了一些答案。IgKI(1,0,0)<ipython-input-53-9b9d0956fa85>:2: RuntimeWarning: overflow encountered in cosh  return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))<ipython-input-53-9b9d0956fa85>:2: RuntimeWarning: overflow encountered in sinh  return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))(0.42102443824067737, 1.3628527263018718e-08)但是當(dāng)我對(duì) (m,n) 使用任何其他值時(shí)，我得到以下結(jié)果：IgKI(1,0,1)<ipython-input-53-9b9d0956fa85>:2: RuntimeWarning: overflow encountered in cosh  return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))<ipython-input-53-9b9d0956fa85>:2: RuntimeWarning: overflow encountered in sinh  return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))<ipython-input-53-9b9d0956fa85>:2: RuntimeWarning: invalid value encountered in double_scalars  return ((np.cosh(x))**m)*((np.sinh(x))**n)*(np.exp(-z*np.cosh(x)))<ipython-input-76-9bee7a5d6456>:2: IntegrationWarning: The occurrence of roundoff error is detected, which prevents   the requested tolerance from being achieved.  The error may be   underestimated.  IntKI = quad(integrand3, 0, np.inf, args=(z,m,n))(nan, nan)那么，我做錯(cuò)了什么？