Good evening! How could one evaluate the following integral 1π∫∞0√x1+xe−xtdx
I have tried the substitution x≡x2 but still I could not manage to get to a final result. Any ideas would be really appreciated!
Also t>0.
Answer
Yes, sub x=u2 to get
2π∫∞0duu21+u2e−tu2=1π∫∞−∞du(1−11+u2)e−tu2=1√πt−1π∫∞−∞due−tu21+u2
The latter integral may be evaluated a few different ways. One way is to multiply and divide by e−t, and differentiate:
1π∫∞−∞due−tu21+u2=etπ∫∞−∞due−t(1+u2)1+u2
and
ddt∫∞−∞due−t(1+u2)1+u2=−e−t∫∞−∞due−tu2=−√πe−t√t
Integrate back with respect to t and get an error function:
∫∞−∞due−t(1+u2)1+u2=−√π∫tdt′e−t′√t′=C−πerf√t
Noting that
∫∞−∞du11+u2=π
We finally have the integral taking the value
1√πt−eterfc√t
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