∫0∞ⅇ-12⁢x2⁢Hen⁡(x)⁢Hen+2⁢m+1⁡(x)⁢sin⁡(x⁢y)⁢ⅆ⁡x=(-1)m⁢12⁢π⁢n!⁢y2⁢m+1⁢ⅇ-12⁢y2⁢Ln(2⁢m+1)⁡(y2).