filestream

filestream.cs

[Route("")]
public HttpResponseMessage GetResult()
{
    var fs = new FileStream(@"C:\data\medline13n0701.xml.gz"), FileMode.Open);
    var response = new HttpResponseMessage(HttpStatusCode.OK)
    {
        Content = new StreamContent(fs)
    };
    response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/xml");
    response.Content.Headers.ContentEncoding.Add("gzip");
    return response;
}