∫e−a22(x+y)(x+y)3/2dy=−2π−−√aerf(a2√x+y−−−−−√)∫e−a22(x+y)(x+y)3/2dy=−2πaerf(a2x+y)∫∞0e−a22(x+y)(x+y)3/2dy=2π−−√aerf(a2x−−√)∫0∞e−a22(x+y)(x+y)3/2dy=2πaerf(a2x)provided R(x)>0∨x∉Rℜ(x)>0∨x∉R Now, consider∫erf(1t√)dt=(t+2)erf(1t√)+2e−1/tt√π√∫erf(1t)dt=(t+2)erf(1t)+2e−1/ttπI=∫b0erf(1t√)dt=(b+2)erf(1b√)+2b√e−1/bπ√−2I=∫0berf(1t)dt=(b+2)erf(1b)+2be−1/bπ−2Now, for large values of bb,I=−2+4b√π√+43πb−−√+O(1b3/2)I=−2+4bπ+43πb+O(1b3/2)which would then make2π−−√a∫b0erf(a2x−−√)=−2π−−√a+4b√+23a21b−−√+O(1b3/2)