# Evaluating the Integral $\int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-(x+y)}}{x+y} dxdy$

I need to find , $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-(x+y)}}{x+y} dxdy$$

My attempt:

I tried doing integration wrt x first then substitute $t = x+y$, $$\int_{0}^{\infty} e^{-y} \left( \int_{0}^{\infty} \frac{e^{-x}}{x+y} dx\right) dy = \int_{0}^{\infty} e^{-y} \left( \int_{t=y}^{\infty} \frac{e^{-(t-y)}}{t} dt\right) = \int_{0}^{\infty} \int_{t=y}^{\infty} \frac{e^{-t}}{t} dt$$

I don't know how to proceed further. Any hint on how should i tackle this kind of problem would be appreciated ...

• Try $$\int_{0}^{\infty} \int_{y=0}^{y=t} \frac{e^{-t}}{t} dy\,dt$$ – Did Apr 5 '17 at 8:55
• Okay. Change of integration it is. Got it. Thank you @Did Answer is $1$. – Raindeer Apr 5 '17 at 8:58