求这个无穷积分

Question

如题所述

Answer 1

sin(tx)/x=∫(0→1)sin(tx)dt
所以，
∫(0→+∞)e^(-x)·sin(tx)/x·dx
=∫(0→+∞)e^(-x)·[∫(0→1)sin(tx)dt]·dx
=∫(0→1)dt∫(0→+∞)e^(-x)·sin(tx)·dx
=∫(0→1)t/(1+t²)dt
=1/2·ln(1+t²) |(0→1)
=1/2·ln2

【附注】内层积分的计算
∫(0→+∞)e^(-x)·sin(tx)·dx
=-∫(0→+∞)sin(tx)·d[e^(-x)]
=-e^(-x)sin(tx) |(0→+∞)
+∫(0→+∞)e^(-x)·tcos(tx)·dx
=0+∫(0→+∞)e^(-x)·tcos(tx)·dx
=-∫(0→+∞)tcos(tx)·d[e^(-x)]
=-e^(-x)·tcos(tx) |(0→+∞)
-∫(0→+∞)e^(-x)·t²sin(tx)·dx
=t-t²∫(0→+∞)e^(-x)·sin(tx)·dx

移项可得：
(1+t²)∫(0→+∞)e^(-x)·sin(tx)·dx=t
∴∫(0→+∞)e^(-x)·sin(tx)·dx=t/(1+t²)追问

谢谢

追答

前面打错了，更正一下
sin(tx)/x=∫(0→t)cos(xy)dy
所以，
∫(0→+∞)e^(-x)·sin(tx)/x·dx
=∫(0→+∞)e^(-x)·[∫(0→t)cos(xy)dy]·dx
=∫(0→t)dy∫(0→+∞)e^(-x)·cos(xy)·dx
=∫(0→t)1/(1+y²)dy
=arctany |(0→t)
=arctant

【附注】内层积分的计算
∫(0→+∞)e^(-x)·cos(xy)·dx
=-∫(0→+∞)cos(xy)·d[e^(-x)]
=-e^(-x)cos(xy) |(0→+∞)
-∫(0→+∞)e^(-x)·ysin(xy)·dx
=1-∫(0→+∞)e^(-x)·ysin(xy)·dx
=1+∫(0→+∞)ysin(xy)·d[e^(-x)]
=1+e^(-x)·ysin(xy) |(0→+∞)
-∫(0→+∞)e^(-x)·y²cos(xy)·dx
=1-y²∫(0→+∞)e^(-x)·cos(xy)·dx

移项可得：
(1+y²)∫(0→+∞)e^(-x)·cos(xy)·dx=1
∴∫(0→+∞)e^(-x)·cos(xy)dx=1/(1+y²)

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